Question

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.

Answer 1

Answer:

The specific heat of the object $C_{obj}$ = 0.457 $\frac{KJ}{kg K}$

Explanation:

Mass of the object $m_{obj}$ = 23.2 gm

Initial temperature $T_{obj}$ = 97 ° c

Mass of the water $m_{w}$ = 90 gm

Initial temperature of water $T_{w}$ = 20.5 ° c

Final temperature of both water & object $T_{f}$ = 22.6 ° c

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

⇒ $m_{obj}$ $C_{obj}$ ( $T_{obj}$ - $T_{f}$ ) =  $m_{w}$ $C_{w}$ ( $T_{f}$ - $T_{w}$)

Put all the values in above formula we get

⇒ 23.2 × $C_{obj}$ ( 97 - 22.6 ) = 90 × 4.18 × ( 22.6 - 20.5 )

⇒ $C_{obj}$ = 0.457 $\frac{KJ}{kg K}$

This is the specific heat of the object.