Answer:
The specific heat of the object [tex]C_{obj}[/tex] = 0.457 [tex]\frac{KJ}{kg K}[/tex]
Explanation:
Mass of the object [tex]m_{obj}[/tex] = 23.2 gm
Initial temperature [tex]T_{obj}[/tex] = 97 ° c
Mass of the water [tex]m_{w}[/tex] = 90 gm
Initial temperature of water [tex]T_{w}[/tex] = 20.5 ° c
Final temperature of both water & object [tex]T_{f}[/tex] = 22.6 ° c
It is given that heat lost by the object = heat gain by the water
⇒ [tex]m_{obj}[/tex] [tex]C_{obj}[/tex] ( [tex]T_{obj}[/tex] - [tex]T_{f}[/tex] ) = [tex]m_{w}[/tex] [tex]C_{w}[/tex] ( [tex]T_{f}[/tex] - [tex]T_{w}[/tex])
Put all the values in above formula we get
⇒ 23.2 × [tex]C_{obj}[/tex] ( 97 - 22.6 ) = 90 × 4.18 × ( 22.6 - 20.5 )
⇒ [tex]C_{obj}[/tex] = 0.457 [tex]\frac{KJ}{kg K}[/tex]
This is the specific heat of the object.
