The specific heat of a 90 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 129
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.
Heat released by water = mass x specific heat capacity of water x temperature change
= 129 * 1 * (80 - 54) = 3354 cal
Heat absorbed by material = mass x specific heat capacity of material x temperature change
= 90 * c * (54 - 20) = 3060c
Heat absorbed by calorimeter = mass x specific heat capacity of Cu x temperature change
= 25 x 0.092 x (54 - 20) = 78.2 cal
Heat absorbed by water = mass x specific heat capacity of water x temperature change
= 60 x 1 x (54 - 20) = 2040 cal
Conservation of energy requires total heat absorbed = total heat released
3060c + 78.2 + 2040 = 3354
c = 0.6477 cal/g.K
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