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## Homework Statement

"Using the above calorimetre with 100.0 g of water of in it at 20.0°C, a 50.00 g sample of a metal at 100.0°C is added. The temperature of the calorimetre rises to 27.2°C. What is the specific heat? (S.H. = 0.20 cal/g°C)"

## Homework Equations

Heat capacity of calorimetre used = 8.0 cal/°C

## The Attempt at a Solution

(27.2 - 20)°C ⋅ 8.0 cal/°C = 57.6 cal

100.0 g ⋅ 7.2°C = 720 calories absorbed by water

50.00 g ⋅ (27.2 - 100) = -3640 calories released by metal

-3640 + 720 = -2920 calories absorbed by calorimetre

(-2920 cal/7.2°C⋅100 g) = 4.05555... cal/g°C ≠ (0.20 cal/g°C)

They already gave me the answer (0.20 cal/g°C) but they want a process, and I am almost sure I am calculating the last step wrong.