Question

A tank contains 2 kg of water at 25°C. Into this tank, a person dropped a 0.9-kg piece of metal that has a temperature of 75°C. What will be the equilibrium temperature of the mixture? Specific heat of metal is 448 J/kg-°C)

Answer 1

Suppose the final temperature is T.

Now Using energy conservation:

Heat gained by 2 kg water = Heat released by 0.9 kg metal

Q1 = Q2

m1*C1*dT1 = m2*C2*dT2

dT1 = Tf - Ti = T - 25

dT2 = 75 - T

m1 = mass of water = 2 kg

m2 = mass of metal = 0.9 kg

C1 = Specific heat capacity of water = 4186 J/kg-°C

C2 = Specific heat capacity of metal = 448 J/kg-°C

So, m1*C1*dT1 = m2*C2*dT2

2*4186*(T - 25) = 0.9*448*(75 - T)

T = [0.9*448*75 + 2*4186*25]/(2*4186 + 0.9*448)

T = 27.3 °C

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