(2007, François G. Amar, all rights reserved)

Heat Transfer within a Calorimeter

Consider the situation in which a hot piece of metal is placed into an insulated container of water. Two observations can be made

1. The metal will cool and the water will heat up (Zeroth Law of Thermodynamics)

2. The heat lost by the metal, q_metal is equal in magnitude to the heat gained by the water, q_water.

However, the signs of the q terms are opposite.

We can use these two observations to calculate the final temperature of the water and the metal once thermal equilibrium has been achieved. Let's first work in terms of algebra with symbols and then we'll put in some numbers.

C _metal is the heat capacity of the metal

C _water is the heat capacity of the water

T_metal is the initial temperature of the metal

T _water is the initial temperature of the water

T_final is the final common temperature of the equilibrium system.

First we can say that q_metal + q _water = 0.

This follows from observation 2 above since the process occurs in an insulated container.

Let us now rewrite these terms

q_metal = C_metal (T _final - T_metal)

q_water= C_water (T _final - T_water)

Note that q_metal is a negative quantity and that q_water is a positive quantity because of observation 1 that

T_metal > T_final> T _water.

Now plug these relations for q in the above expression:

C_metal (T _final - T_metal) + C_waterr (T _final - T_water) = 0

Rearrange:

(C _metal + C_water) T _final - (C _metal T _metal + C _water T_water) = 0

Now let us put in some numbers:

55.0 g of Fe with specific heat 0.450 J/g°C initially at 650 °C

970 g of water with specific heat 4.18 J/g-°C initially at 22.0 °C

C_metal = (0.450 J/g°C)( 55.0 g) = 24.8 J/°C

C_water = (4.18 J/g°C)( 970. g) = 4054.6 J/°C (keeping an extra sig fig till the end)