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A metal object with mass of 20.4 g is heated to 97.0 °C and then transferred to an insulated container containing 80.9 g of water at 20.5 °C. The water temperature rises and the temperature of the metal object falls until they both reach the same final temperature of 22.7 °C. What is the specific heat of this metal object? Assume that all the heat lost by the metal object is absorbed by the water.

Respuesta :

Answer:

Explanation:

Heat lost by metal object = mass x specific heat x fall in temperature

= 20.4 x S x ( 97 - 22.7 )

= 1515.72 S

Heat gained by water

= 80.9 x 1 x ( 22.7 - 20.5 )

= 177.98 Cals

heat lost = heat gained

1515.72 S = 177.98

S = .117 Cals /g . C

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