Question

If a piece of Cd with a mass = 37.6 g and initial temperature of 100.0 o C is dropped in 25.0 mL
of water at 23 o C, what is the final temperature of the system? specific heat of Cd = 0.232
J/g o C; specific heat of water = 4.184 J/g o C.

Answer 1

Let us denote water by symbol 1 and Cd by symbol 2

m1 = 25.0 g (since density of water is 1 g/mL and volume is 25.0 mL)

T1 = 23.0 oC

C1 = 4.184 J/goC

m2 = 37.6 g

T2 = 100.0 oC

C2 = 0.232 J/goC

T = to be calculated

Let the final temperature be T oC

we have below equation to be used:

heat lost by 2 = heat gained by 1

m2*C2*(T2-T) = m1*C1*(T-T1)

37.6*0.232*(100.0-T) = 25.0*4.184*(T-23.0)

8.7232*(100.0-T) = 104.6*(T-23.0)

872.32 - 8.7232*T = 104.6*T - 2405.8

T= 28.9 oC

Answer: 28.9 oC