A coffee-cup calorimeter contains 150.0g of water at 25.1C. A 121g block of copp

Question

A coffee-cup calorimeter contains 150.0g of water at 25.1C. A 121g block of copper metal is heated to 100.4C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g*K. The Cu is added to the Calorimeter, and after a time the contents of the cup reach a constant temperature of 30.3C.

I have already calculated the amout of heat lost by the copper block- 3270J
And the heat gained by the water- 3260J

The difference between these answers is due to the heat lost through the styrofoam.

How, now, do I calculate the heat capacity of the calorimeter?

Explanation / Answer

the amout of heat lost and the heat gained by the water is not 3270J and 3260J

amount of heat lost by copper = 121.0*(100.4-30.0)*0.385 = 3279.58 ~ 3280J

& amount of heat gained by water = 150.0*(30.0-25.1)*4.18 = 3072.3 ~ 3072J

heat capacity of the calorimeter = (3280-3072)/(30.0-25.1) = 42.45 J/K

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