A metal object with mass of 23.2 g 23.2 g is heated to 97.0 °C 97.0 °C and then transferred to an insulated container containing 90.0 g 90.0 g of water at 20.5 °C. 20.5 °C. The water temperature rises and the temperature of the metal object falls until they both reach the same final temperature of 22.6 °C. 22.6 °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.

The specific heat of the object = 0.457

Explanation:

Mass of the object = 23.2 gm

Initial temperature = 97 ° c

Mass of the water = 90 gm

Initial temperature of water = 20.5 ° c

Final temperature of both water & object = 22.6 ° c

It is given that heat lost by the object = heat gain by the water

⇒ ( - ) =   ( - )

Put all the values in above formula we get

⇒ 23.2 × ( 97 - 22.6 ) = 90 × 4.18 × ( 22.6 - 20.5 )

⇒ = 0.457

This is the specific heat of the object.

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