The specific heat of a 101 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 125 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.

c = 0.329 cal/g. K

Explanation:

total Heat released = mass x specific heat capacity of water x temperature change

= 125 x 1 x (80 - 54) = 3250 cal

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

= 101 x c x (54 - 20) = 3434*c cal

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

3434*c + 78.2 + 2040 = 3250

c = 0.329 cal/g. K