Question

The specific heat of a 103 g block of material is to be determined. The block is placed in a 25 g copper calorimeter that also holds 60 g of water. The system is initially at 20°C. Then 122 mL of water at 80°C are added to the calorimeter vessel. When thermal equilibrium is attained, the temperature of the water is 54°C. Determine the specific heat of the block.

Answer 1

Heat released by water = mass * specific heat capacity of water * temperature change

= 122 * 1 * (80 - 54) = 3172 cal

Heat absorbed by material = mass * specific heat capacity of material * temperature change

= 103 * c * (54 - 20) = 3502c

Heat absorbed by calorimeter = mass * specific heat capacity of Cu * temperature change

= 25 * 0.092 * (54 - 20) = 78.2 cal

Heat absorbed by water = mass * specific heat capacity of water * temperature change

= 60 * 1 * (54 - 20) = 2040 cal

Conservation of energy requires total heat absorbed = total heat released

3502c + 78.2 + 2040 = 3172

c = 0.301 cal/g.K