Question

A metal piece of 50g specific heat 0.6cal/g*C initially at 120*C is dropped in 1.6 kg of water at 25*C . Find the final temperature of the mixture.

Answer 1

Final Temperature of mixture will be 26.7° C.

Temperature of metal piece will be equal to temperature of water.

M1c1(T1-Tf) = M2C2(Tf-T)

M1 is the mass of metall piece

C1 is the specific heat of metal piece

T1 is initial temperature of metal piece

Tf is final Temperature of the mixture

M2 is the mass of water

C2 is specific heat of water

= 1 cal/g °C

T2 is the temperature of water

Putting the given values in equation

50*0.6*(120-Tf) = 1.6*1000*1*(Tf-25)

30*(120-Tf) = 1600*(Tf-25)

3600-30Tf = 1600Tf - 40000

40000+3600 = 1600Tf + 30Tf

43600 = 1630 Tf

Tf = 26.7 °C

Therefore, final Temperature of the mixture will be 26.7°C.