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In general, specific heat also depends on temperature. Thus, a precise definition of c for a substance must be given in terms of an infinitesimal change in temperature. To do this, we note that c = 1 m Δ Q Δ T and replace Δ with d :

c = 1 m d Q d T .

Except for gases, the temperature and volume dependence of the specific heat of most substances is weak at normal temperatures. Therefore, we will generally take specific heats to be constant at the values given in the table.

Calculating the required heat

A 0.500-kg aluminum pan on a stove and 0.250 L of water in it are heated from 20.0 ° C to 80.0 ° C . (a) How much heat is required? What percentage of the heat is used to raise the temperature of (b) the pan and (c) the water?

Strategy

We can assume that the pan and the water are always at the same temperature. When you put the pan on the stove, the temperature of the water and that of the pan are increased by the same amount. We use 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 [link] .

Solution

  1. Calculate the temperature difference:
    Δ T = T f T i = 60.0 ° C .
  2. Calculate the mass of water. Because the density of water is 1000 kg/m 3 , 1 L of water has a mass of 1 kg, and the mass of 0.250 L of water is m w = 0.250 kg .
  3. Calculate the heat transferred to the water. Use the specific heat of water in [link] :
    Q w = m w c w Δ T = ( 0.250 kg ) ( 4186 J/kg ° C ) ( 60.0 ° C ) = 62.8 kJ .
  4. Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in [link] :
    Q Al = m A1 c A1 Δ T = ( 0.500 kg ) ( 900 J/kg ° C ) ( 60.0 ° C ) = 27.0 kJ .
  5. Find the total transferred heat:
    Q Total = Q W + Q Al = 89.8 kJ .

Significance

In this example, the heat transferred to the container is a significant fraction of the total transferred heat. Although the mass of the pan is twice that of the water, the specific heat of water is over four times that of aluminum. Therefore, it takes a bit more than twice as much heat to achieve the given temperature change for the water as for the aluminum pan.

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[link] illustrates a temperature rise caused by doing work. (The result is the same as if the same amount of energy had been added with a blowtorch instead of mechanically.)

Calculating the temperature increase from the work done on a substance

Truck brakes used to control speed on a downhill run do work, converting gravitational potential energy into increased internal energy (higher temperature) of the brake material ( [link] ). This conversion prevents the gravitational potential energy from being converted into kinetic energy of the truck. Since the mass of the truck is much greater than that of the brake material absorbing the energy, the temperature increase may occur too fast for sufficient heat to transfer from the brakes to the environment; in other words, the brakes may overheat.

Figure shows a truck on a road. There is smoke near its tires.
The smoking brakes on a braking truck are visible evidence of the mechanical equivalent of heat.

Calculate the temperature increase of 10 kg of brake material with an average specific heat of 800 J/kg · °C if the material retains 10% of the energy from a 10,000-kg truck descending 75.0 m (in vertical displacement) at a constant speed.

Practice Key Terms 7

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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