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In an experiment, 25.5 g of metal was heated to 98.0°C and then quickly transferred to...

In an experiment, 25.5 g of metal was heated to 98.0°C and then quickly transferred to 150.0 g of water in a calorimeter. The initial temperature of the water was 21.0°C, and the final temperature after the addition of the metal was 32.5°C. Assume the calorimeter behaves ideally and does not absorb or release heat.

What is the value of the specific heat capacity (in J/g•°C) of the metal?

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Answer #1

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

m1 = 150.0 g

T1 = 21.0 oC

C1 = 4.184 J/goC

m2 = 25.5 g

T2 = 98.0 oC

C2 = to be calculated

T = 32.5 oC

we have below equation to be used:

heat lost by 2 = heat gained by 1

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

25.5*C2*(98.0-32.5) = 150.0*4.184*(32.5-21.0)

1670.25*C2 = 7217.4

C2= 4.3211 J/goC

Answer: 4.32 J/goC

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