Question

A 125-g block of aluminum at room temperature (21°C) is placed into 150 g of boiling water (100°C). If the aluminum block and water form a closed system, what will their temperature be when they achieve thermal equilibrium?

Answer 1

Suppose the final temperature is T.

Now Using energy conservation:

Heat gained by aluminium = Heat released by water

Q1 = Q2

m1*C1*dT1 = m2*C2*dT2

dT1 = Tf - Ti = T - Initial temperature of aluminium = T - 21

dT2 = Initial temperature of water - T = 100 - T

m1 = mass of aluminium = 125 g = 0.125 kg

m2 = mass of water = 150 g = 0.150 kg

C1 = Specific Heat capacity of aluminium = 900 J/kg-C

C2 = Specific Heat capacity of water = 4186 J/kg-C

Now using given values:

0.125*900*(T - 21) = 0.150*4186*(100 - T)

Now Solving above equation

T = (0.150*4186*100 + 0.125*900*21)/(0.125*900 + 0.150*4186) = 87.996 degC

T = 88 degC = Final temperature of system

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