Answer to Question #99578 in General Chemistry for Beverlie

Question

In an experiment, 28.5 g of metal was heated to 98.0°C and then quickly transferred to 150.0 g of water in a calorimeter. The initial temperature of the water was 25.0°C, and the final temperature after the addition of the metal was 32.5°C. Assume the calorimeter behaves ideally and does not absorb or release heat.

What is the value of the specific heat capacity (in J/g•°C) of the metal?

Answer 1

Heat lost by metal = heat gained by water

$Q_{lost} =Q_{gained}$

$Q_{lost} = cm(T_2-T_1)=c\times 28.5 (32.5-98) = -1866.75 \times c$

$Q_{gained} = cm(T_2-T_1) = 4.186\times 150\times(32.5 - 25) = 4709.25 J$

$Q_{lost} = -4709.25 J$

$$-4709.25 = -1866.75\times c$$

$$c = 2.52 J/g\times ^\circ C$$