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

The specific heat of the object
`C_(obj)` = 0.457
`(KJ)/(kg K)`

Step-by-step 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
`(KJ)/(kg K)`

This is the specific heat of the object.