Question

The temperature of a piece of silver (specific heat Sag = 0.031j/g.°C) with a mass of 362 g decreased by 58 °C when it was added to a 98.56 g sample of water (specific heat of water Sw = 4.184 j/g °C) in a constant pressure calorimeter of negligible heat capacity. What is the final temperature of water if its initial temperature was 23.4 °C ?

Answer 1

heat lost by hot silver = heat gained by water

m1*s1*DT1 = m2*s2*DT2

m1 = mass of hot silver = 362 g

S1 = specific hot silver = 0.031 j/g.c

DT1 = 58

m2 = mass of water = 98.56 g

S = specific heat of water = 4.184 j/g.c

DT2 = x-23.4

362*0.031*58 = 98.56*4.184*(x-23.4)

x = 24.98

x = final temperature of water = 24.98 c