Question

A 94.7-g sample of silver (s = 0.237 j/(g · °c)), initially at 348.25°c, is added to an insulated vessel containing 143.6 g of water (s = 4.18 j/(g · °c)), initially at 13.97°c. at equilibrium, the final temperature of the metal–water mixture is 22.63°c. how much heat was absorbed by the water? the heat capacity of the vessel is 0.244 kj/°c.

Answer 1

Q = m*C*dT

m = mass

C = specific heat capacity

dT = Temperature change

Q = heat evolved

Heat lost by metal is equal to the sum of heat taken up by water and vessel. Since water is taken in the vessel so

the initial and final temperatures would be same for vessel and water...

Q = 94.7 *0.237 *(22.63 - 348.25) = 7308.18 Joules = 7.3 KJ

Heat lost by silver = 7.3 KJ

Heat absorbed by water = 143.6*4.18*(22.63 - 13.97) = 5.2 KJ