The Specific heat of a 96.2 g block of material is to be determined. The block i

Question

The Specific heat of a 96.2 g block of material is to be determined. The block is placed in a 18.1 g copper calorimeter that also holds 61.9 g of water. The system is initially at 24.5C, then 110.1 g of water at 80.9C are added to the calorimeter vessel. When thermal equilibrium is attained, the temperature of the water is 51.5C.
Determine the specificheat of the block. The secific heat of copper and water are, 0.386 and 4.183.

This is the only problem that I currently am struggling with to solve on my homework..

Explanation / Answer

We use the law of calorimetry which states that heat lost = heat gained by the system. So, Let specific heat of object = c 96.2*c(51.5-24.5) + 18.1*.386*(51.5-24.5) + 61.9 * 4.183 *(51.5-24.5) = 110.1*4.183*(80.9-51.5) 96.2*c(51.5-24.5) = (110.1*4.183*(80.9-51.5)) - (18.1*.386*(51.5-24.5)) - (61.9 * 4.183*(51.5-24.5)) c = 6360.43392/(96.2*(51.5-24.5)) = 2.448769 I hope this answers you!

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