Question

A silver block, initially at 59.3 ∘C, is submerged into 100.0 g of water at 25.3 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 26.5 ∘C.

What is the mass of the silver block?

Answer 1

we calculate the temperature change of the water = 26.5 - 25.3 = 1.2 deg C

then calculate the temperature change of the silver = 59.3 - 26.5 = 32.8 deg C

There is nothing to relate mass to temperature change. To do more with this information you need the specific heats of silver and water and you assume the energy gained by the water is lost from the silver so you can now relate the mass to the delta T (change in temperature).

we know, specific heat of water = 1 cal/g deg C = 4.186 J/ g deg C

specific heat Ag = 0.057 cal/g deg C = 0.239 J/g deg C

heat gained by water = heat lost by Ag = 4.186 J/g deg C * 100.0 g * 1.2 deg C = 502.32 joules

mass Ag = 502.32 J / (0.239 J/g deg C * 32.8 deg C ) = 64.07 g Ag