if 40 kcal of heat is added to 2.0 kg of water, what is the resulting temperature change?

Learning Objectives

By the stop of this section, you will be able to:

• Observe estrus transfer and change in temperature and mass.
• Calculate terminal temperature later heat transfer between two objects.

1 of the major furnishings of heat transfer is temperature change: heating increases the temperature while cooling decreases it. We assume that there is no phase alter and that no work is done on or by the organization. Experiments bear witness that the transferred oestrus depends on three factors—the modify in temperature, the mass of the organisation, and the substance and phase of the substance.

Figure 1. The heat Q transferred to cause a temperature change depends on the magnitude of the temperature modify, the mass of the system, and the substance and stage involved. (a) The corporeality of heat transferred is directly proportional to the temperature change. To double the temperature change of a mass k, y'all need to add twice the heat. (b) The corporeality of heat transferred is also direct proportional to the mass. To crusade an equivalent temperature modify in a doubled mass, you lot need to add together twice the oestrus. (c) The amount of heat transferred depends on the substance and its phase. If it takes an amount Q of rut to cause a temperature change ΔT in a given mass of copper, it will take 10.8 times that amount of heat to crusade the equivalent temperature change in the same mass of water bold no phase modify in either substance.

The dependence on temperature change and mass are easily understood. Owing to the fact that the (average) kinetic energy of an atom or molecule is proportional to the absolute temperature, the internal energy of a system is proportional to the accented temperature and the number of atoms or molecules. Attributable to the fact that the transferred heat is equal to the alter in the internal energy, the oestrus is proportional to the mass of the substance and the temperature modify. The transferred heat also depends on the substance so that, for case, the heat necessary to enhance the temperature is less for alcohol than for h2o. For the same substance, the transferred heat too depends on the phase (gas, liquid, or solid).

Heat Transfer and Temperature Change

The quantitative human relationship between heat transfer and temperature change contains all three factors:Q =mcΔT, where Q is the symbol for rut transfer, m is the mass of the substance, and ΔT is the change in temperature. The symbol c stands for specific rut and depends on the material and phase. The specific estrus is the amount of heat necessary to change the temperature of 1.00 kg of mass by one.00ºC. The specific heat c is a property of the substance; its SI unit of measurement is J/(kg ⋅ K) or J/(kg ⋅ ºC). Recall that the temperature change (ΔT) is the same in units of kelvin and degrees Celsius. If heat transfer is measured in kilocalories, then the unit of measurement of specific heat is kcal/(kg ⋅ ºC).

Values of specific heat must generally be looked up in tables, because there is no simple way to summate them. In full general, the specific estrus also depends on the temperature. Table 1 lists representative values of specific estrus for various substances. Except for gases, the temperature and book dependence of the specific oestrus of near substances is weak. We see from this table that the specific estrus of water is five times that of glass and ten times that of iron, which means that it takes v times as much heat to raise the temperature of water the same amount as for glass and ten times as much heat to raise the temperature of h2o as for iron. In fact, water has 1 of the largest specific heats of any material, which is important for sustaining life on Globe.

Instance one. Calculating the Required Estrus: Heating Water in an Aluminum Pan

A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from twenty.0ºC to 80.0ºC. (a) How much heat is required? What percentage of the heat is used to raise the temperature of (b) the pan and (c) the h2o?

Strategy

The pan and the water are always at the same temperature. When you put the pan on the stove, the temperature of the h2o and the pan is increased past the same corporeality. We utilise the equation for the heat transfer for the given temperature change and mass of water and aluminum. The specific heat values for water and aluminum are given in Table ane.

Solution

Considering water is in thermal contact with the aluminum, the pan and the water are at the aforementioned temperature.

Summate the temperature difference:

ΔT = T _f − T _i = 60.0ºC.

Calculate the mass of water. Because the density of water is k kg/m³, one liter of h2o has a mass of 1 kg, and the mass of 0.250 liters of water is grand _w  = 0.250 kg.

Calculate the rut transferred to the h2o. Use the specific heat of h2o in Table 1:

Q _w =m _w c _{due west}ΔT = (0.250 kg)(4186 J/kgºC)(60.0ºC) = 62.8 kJ.

Calculate the estrus transferred to the aluminum. Use the specific heat for aluminum in Table 1:

Q _Al =m _Al c _AlΔT= (0.500 kg)(900 J/kgºC)(threescore.0ºC) = 27.0 × ten⁴ J = 27.0 kJ.<

Compare the percentage of heat going into the pan versus that going into the water. First, find the total transferred estrus:

Q _Total =Q _westward + Q _Al= 62.viii kJ + 27.0 kJ = 89.8 kJ.

Thus, the amount of heat going into heating the pan is

[latex]\frac{27.0\text{ kJ}}{89.eight\text{ kJ}}\times100\%=30.one\%\\[/latex]

and the amount going into heating the water is

[latex]\frac{62.viii\text{ kJ}}{89.8\text{ kJ}}\times100\%=69.9\%\\[/latex].

Discussion

In this example, the heat transferred to the container is a significant fraction of the total transferred rut. Although the mass of the pan is twice that of the water, the specific oestrus of h2o is over 4 times greater than that of aluminum. Therefore, it takes a chip more than twice the heat to accomplish the given temperature change for the water as compared to the aluminum pan.

Example 2. Calculating the Temperature Increment from the Work Done on a Substance: Truck Brakes Overheat on Downhill Runs

Effigy two. The smoking brakes on this truck are a visible testify of the mechanical equivalent of heat.

Truck brakes used to control speed on a downhill run do work, converting gravitational potential energy into increased internal free energy (higher temperature) of the brake material. This conversion prevents the gravitational potential energy from being converted into kinetic free energy of the truck. The problem is that the mass of the truck is big compared with that of the restriction fabric arresting the energy, and the temperature increase may occur too fast for sufficient heat to transfer from the brakes to the environment.

Summate the temperature increase of 100 kg of brake textile with an average specific heat of 800 J/kg ⋅ ºC if the fabric retains 10% of the energy from a x,000-kg truck descending 75.0 m (in vertical deportation) at a constant speed.

Strategy

If the brakes are not applied, gravitational potential energy is converted into kinetic energy. When brakes are applied, gravitational potential energy is converted into internal free energy of the restriction material. We first calculate the gravitational potential energy (Mgh) that the entire truck loses in its descent and and then find the temperature increase produced in the brake fabric alone.

Solution

1. Calculate the change in gravitational potential energy as the truck goes downhillMgh = (10,000 kg)(9.eighty m/sⁱⁱ)(75.0 k) = 7.35 × x⁶ J.
2. Summate the temperature from the heat transferred using Q =Mgh and [latex]\Delta{T}=\frac{Q}{mc}\\[/latex], where yard is the mass of the brake material. Insert the values m= 100 kg and c= 800 J/kg ⋅ ºC to find [latex]\Delta{T}=\frac{\left(vii.35\times10^6\text{ J}\right)}{\left(100\text{ kg}\right)\left(800\text{ J/kg}^{\circ}\text{C}\correct)}=92^{\circ}C\\[/latex].

Discussion

This temperature is close to the boiling point of water. If the truck had been traveling for some time, then merely before the descent, the restriction temperature would likely be college than the ambient temperature. The temperature increase in the descent would probable enhance the temperature of the restriction material to a higher place the boiling indicate of water, so this technique is not practical. Yet, the aforementioned idea underlies the recent hybrid technology of cars, where mechanical energy (gravitational potential energy) is converted past the brakes into electric energy (bombardment).

Table 1. Specific Heats[1] of Various Substances
Substances	Specific oestrus (c)
Solids	J/kg ⋅ ºC	kcal/kg ⋅ ºC[two]
Aluminum	900	0.215
Asbestos	800	0.19
Physical, granite (boilerplate)	840	0.twenty
Copper	387	0.0924
Glass	840	0.twenty
Gold	129	0.0308
Man body (boilerplate at 37 °C)	3500	0.83
Ice (boilerplate, −50°C to 0°C)	2090	0.fifty
Atomic number 26, steel	452	0.108
Lead	128	0.0305
Argent	235	0.0562
Forest	1700	0.4
Liquids
Benzene	1740	0.415
Ethanol	2450	0.586
Glycerin	2410	0.576
Mercury	139	0.0333
Water (15.0 °C)	4186	1.000
Gases [3]
Air (dry)	721 (1015)	0.172 (0.242)
Ammonia	1670 (2190)	0.399 (0.523)
Carbon dioxide	638 (833)	0.152 (0.199)
Nitrogen	739 (1040)	0.177 (0.248)
Oxygen	651 (913)	0.156 (0.218)
Steam (100°C)	1520 (2020)	0.363 (0.482)

Notation that Example 2 is an illustration of the mechanical equivalent of heat. Alternatively, the temperature increment could be produced by a blow torch instead of mechanically.

Instance 3. Calculating the Final Temperature When Oestrus Is Transferred Between Ii Bodies: Pouring Cold H2o in a Hot Pan

Suppose you pour 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan off the stove with a temperature of 150ºC. Assume that the pan is placed on an insulated pad and that a negligible amount of water boils off. What is the temperature when the water and pan reach thermal equilibrium a short time later on?

Strategy

The pan is placed on an insulated pad and then that picayune heat transfer occurs with the surroundings. Originally the pan and water are non in thermal equilibrium: the pan is at a college temperature than the water. Heat transfer and then restores thermal equilibrium once the h2o and pan are in contact. Because heat transfer betwixt the pan and water takes place chop-chop, the mass of evaporated water is negligible and the magnitude of the heat lost by the pan is equal to the oestrus gained by the water. The exchange of heat stops in one case a thermal equilibrium between the pan and the water is achieved. The heat exchange tin can be written as |Q _hot|=Q _cold.

Solution

Utilize the equation for heat transfer Q =mcΔT to express the heat lost by the aluminum pan in terms of the mass of the pan, the specific estrus of aluminum, the initial temperature of the pan, and the last temperature:Q _hot =grand _Al c _Al(T _f − 150ºC).

Express the rut gained by the water in terms of the mass of the water, the specific heat of h2o, the initial temperature of the water and the final temperature:Q _{common cold}=g _{Due west} c _W(T _f − 20.0ºC).

Annotation that Q _hot<0 and Q _{common cold}>0 and that they must sum to naught because the heat lost past the hot pan must be the same equally the oestrus gained past the cold water:

[latex]\brainstorm{array}{lll}Q_{\text{common cold}}+Q_{\text{hot}}&=&0\\Q_{\text{cold}}&=&-Q_{\text{hot}}\\m_{\text{W}}c_{\text{West}}\left(T_{\text{f}}-20.0^{\circ}\text{C}\right)&=&-m_{\text{Al}}c_{\text{Al}}\left(T_{\text{f}}-150^{\circ}\text{C}\right)\finish{array}\\[/latex]

This an equation for the unknown last temperature, T _f.

Bring all terms involving T _f on the left hand side and all other terms on the right hand side. Solve for T _f,

[latex]\displaystyle{T_{\text{f}}}=\frac{m_{\text{Al}}c_{\text{Al}}\left(T_{\text{f}}-150^{\circ}\text{C}\correct)+m_{\text{W}}c_{\text{W}}\left(T_{\text{f}}-20.0^{\circ}\text{C}\right)}{m_{\text{Al}}c_{\text{Al}}+m_{\text{W}}c_{\text{W}}}\\[/latex],

and insert the numerical values:

[latex]\begin{array}{lll}T_{\text{f}}&=&\frac{\left(0.500\text{ kg}\right)\left(900\text{ J/kg}^{\circ}\text{C}\right)\left(150^{\circ}\text{C}\right)+\left(0.250\text{ kg}\right)\left(4186\text{ J/kg}^{\circ}\text{C}\right)\left(xx.0^{\circ}\text{C}\right)}{\left(0.500\text{ kg}\right)\left(900\text{ J/kg}^{\circ}\text{C}\right)+\left(0.250\text{ kg}\right)\left(4186\text{ J/kg}^{\circ}\text{C}\right)}\\\text{ }&=&\frac{88430\text{ J}}{1496.5\text{ J}/^{\circ}\text{C}}\\\text{ }&=&59.1^{\circ}\text{C}\terminate{array}\\[/latex]

Discussion

This is a typical calorimetry problem—2 bodies at different temperatures are brought in contact with each other and exchange rut until a common temperature is reached. Why is the final temperature then much closer to 20.0ºC than 150ºC? The reason is that water has a greater specific heat than most common substances and thus undergoes a pocket-sized temperature change for a given estrus transfer. A big sea, such equally a lake, requires a large amount of heat to increase its temperature appreciably. This explains why the temperature of a lake stays relatively constant during a twenty-four hours even when the temperature modify of the air is large. However, the water temperature does change over longer times (eastward.g., summertime to wintertime).

Accept-Home Experiment: Temperature Change of Land and Water

What heats faster, land or h2o?

To written report differences in rut capacity:

• Place equal masses of dry sand (or soil) and water at the same temperature into two small jars. (The average density of soil or sand is about ane.half-dozen times that of water, so you tin achieve approximately equal masses by using 50% more water by volume.)
• Heat both (using an oven or a heat lamp) for the same amount of time.
• Record the final temperature of the two masses.
• At present bring both jars to the same temperature by heating for a longer period of fourth dimension.
• Remove the jars from the oestrus source and measure out their temperature every 5 minutes for about 30 minutes.

Which sample cools off the fastest? This activity replicates the phenomena responsible for land breezes and sea breezes.

Check Your Agreement

If 25 kJ is necessary to enhance the temperature of a cake from 25ºC to 30ºC, how much heat is necessary to heat the cake from 45ºC to 50ºC?

Solution

The heat transfer depends just on the temperature difference. Since the temperature differences are the same in both cases, the same 25 kJ is necessary in the 2nd case.

Section Summary

• The transfer of heat Q that leads to a change ΔT in the temperature of a body with mass m is Q =mcΔT, where c is the specific oestrus of the textile. This relationship can also exist considered as the definition of specific estrus.

Conceptual Questions

1. What 3 factors touch the rut transfer that is necessary to alter an object's temperature?
2. The brakes in a car increase in temperature by ΔT when bringing the car to residue from a speed 5. How much greater would ΔT be if the car initially had twice the speed? You may assume the car to stop sufficiently fast and so that no heat transfers out of the brakes.

Bug & Exercises

1. On a hot day, the temperature of an fourscore,000-L pond pool increases by 1.50ºC. What is the net heat transfer during this heating? Ignore any complications, such as loss of h2o by evaporation.
2. Prove that ane cal/k · ºC =i kcal/kg · ºC.
3. To sterilize a 50.0-k drinking glass baby bottle, we must raise its temperature from 22.0ºC to 95.0ºC. How much oestrus transfer is required?
4. The aforementioned heat transfer into identical masses of dissimilar substances produces different temperature changes. Calculate the concluding temperature when 1.00 kcal of heat transfers into 1.00 kg of the following, originally at 20.0ºC: (a) water; (b) concrete; (c) steel; and (d) mercury.
5. Rubbing your hands together warms them by converting piece of work into thermal energy. If a woman rubs her hands dorsum and forth for a total of xx rubs, at a distance of seven.fifty cm per rub, and with an average frictional force of 40.0 Northward, what is the temperature increase? The mass of tissues warmed is but 0.100 kg, by and large in the palms and fingers.
6. A 0.250-kg block of a pure textile is heated from 20.0ºC to 65.0ºC past the addition of 4.35 kJ of free energy. Calculate its specific heat and identify the substance of which it is most likely composed.
7. Suppose identical amounts of heat transfer into different masses of copper and water, causing identical changes in temperature. What is the ratio of the mass of copper to water?
8. (a) The number of kilocalories in nutrient is determined past calorimetry techniques in which the food is burned and the amount of oestrus transfer is measured. How many kilocalories per gram are in that location in a v.00-g peanut if the energy from burning it is transferred to 0.500 kg of water held in a 0.100-kg aluminum loving cup, causing a 54.9ºC temperature increase? (b) Compare your respond to labeling data plant on a package of peanuts and comment on whether the values are consistent.
9. Post-obit vigorous exercise, the body temperature of an 80.0-kg person is xl.0ºC. At what rate in watts must the person transfer thermal energy to reduce the the body temperature to 37.0ºC in 30.0 min, assuming the body continues to produce energy at the charge per unit of 150 W? i watt = one joule/2nd or 1 W = one J/due south.
10. Even when shut downwardly after a flow of normal utilise, a big commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive disuse of fission products. This heat transfer causes a rapid increment in temperature if the cooling arrangement fails (i watt = ane joule/second or 1 W = 1 J/s and ane MW = i megawatt). (a) Summate the rate of temperature increase in degrees Celsius per second (ºC/south) if the mass of the reactor core is 1.sixty × 10⁵ kg and it has an boilerplate specific heat of 0.3349 kJ/kg ⋅ ºC. (b) How long would it have to obtain a temperature increment of 2000ºC, which could cause some metals holding the radioactive materials to melt? (The initial rate of temperature increase would be greater than that calculated here because the heat transfer is concentrated in a smaller mass. Later, however, the temperature increase would slow downward because the 5 × 10⁵-kg steel containment vessel would also begin to estrus up.)

Figure iii. Radioactive spent-fuel pool at a nuclear power plant. Spent fuel stays hot for a long time. (credit: U.S. Department of Energy)

Glossary

specific heat: the amount of heat necessary to change the temperature of 1.00 kg of a substance by 1.00 ºC

Selected Solutions to Problems & Exercises

i. 5.02 × 10⁸ J

3. 3.07 × 10ⁱⁱⁱ J

five. 0.171ºC

7. 10.8

9. 617 W

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Source: https://courses.lumenlearning.com/physics/chapter/14-2-temperature-change-and-heat-capacity/

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