Question

A hot lump of 26.3 g of aluminum at an initial temperature of 67.2 °C is placed in 50.0 mL of H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water given that the specific heat of aluminum is 0.903 J/(g·°C)? Assume no heat is lost to surroundings.

Answer 1

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

m1 = 50.0 g (since density of water is 1 g/mL and volume here is 50.0 mL)

T1 = 25.0 oC

C1 = 4.184 J/goC

m2 = 26.3 g

T2 = 67.2 oC

C2 = 0.903 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)

26.3*0.903*(67.2-T) = 50.0*4.184*(T-25.0)

23.7489*(67.2-T) = 209.2*(T-25.0)

1595.9261 - 23.7489*T = 209.2*T - 5230

T= 29.3 oC

Answer: 29.3 oC