Which Of The Following Temperature Changes Are Equivalent

Example 1. Calculating the Required Heat: 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 20.0ºC 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 water?

Strategy

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 the pan is 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 Table 1.

Solution

Because water is in thermal contact with the aluminum, the pan and the water are at the same temperature.

Calculate the temperature difference:

ΔT = Tf − Ti = 60.0ºC.

Calculate the mass of water. Because the density of water is 1000 kg/m3, one liter of water has a mass of 1 kg, and the mass of 0.250 liters of water is mw = 0.250 kg.

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

Qw = mwcwΔT = (0.250 kg)(4186 J/kgºC)(60.0ºC) = 62.8 kJ.

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

QAl = mAlcAlΔT = (0.500 kg)(900 J/kgºC)(60.0ºC) = 27.0 × 104 J = 27.0 kJ.<

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

QTotal = Qw + QAl = 62.8 kJ + 27.0 kJ = 89.8 kJ.

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

[latex]frac{27.0text{ kJ}}{89.8text{ kJ}}times100%=30.1%[/latex]

and the amount going into heating the water is

[latex]frac{62.8text{ kJ}}{89.8text{ kJ}}times100%=69.9%[/latex].

Discussion

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 greater than that of aluminum. Therefore, it takes a bit more than twice the heat to achieve the given temperature change for the water as compared to the aluminum pan.

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