A 25.5-g aluminum block is warmed to 65.1C and plunged into an insulated beaker

Question

A 25.5-g aluminum block is warmed to 65.1C and plunged into an insulated beaker containing 55.0 g of water initially at 22.1 C. The aluminum and water are allowed to come to thermal equilibrium. Specific Heat of H2O= 4.19 J/g•C, Specific Hear of Al = 0.903 J/g •C. Assuming that no heat is lost, what is the final temperature of the water and aluminum? A 25.5-g aluminum block is warmed to 65.1C and plunged into an insulated beaker containing 55.0 g of water initially at 22.1 C. The aluminum and water are allowed to come to thermal equilibrium. Specific Heat of H2O= 4.19 J/g•C, Specific Hear of Al = 0.903 J/g •C. Assuming that no heat is lost, what is the final temperature of the water and aluminum?

Explanation / Answer

Let substance 1 be water and 2 be Al

Given:

m1 = 55.0 g

T1 = 22.1 oC

C1 = 4.19 J/goC

m2 = 25.5 g

T2 = 65.1 oC

C2 = 0.903 J/goC

T = to be calculated

Let the final temperature be T oC

use:

heat lost by 2 = heat gained by 1

m2*C2*(T2-T) = m1*C1*(T-T1)

25.5*0.903*(65.1-T) = 55.0*4.19*(T-22.1)

T= 26.0 oC

Answer: 26.0 oC

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