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.
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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.
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