A mass of 50 g of a certain metal at `150^@`C is immersed in 100 g of water at `11^@`C. The final temperature is `20^@`C. Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is 4.2 `g^(-1)K^(-1)`

Explanation

Given : For solid (Hot body) : `m_(1)=50g,T_(1)=150^(@)C,T=20^(@)C`
Fall in temperature of solid,
`DeltaT_(1)=(150-20)^(@)C`
= `130^(@)C`
`c_(1)=?`
For water (Cold body) : `m_(2)=100g,T_(2)=11^(@)C`,T=20^(@)C`
Rise in temperature of water,
`DeltaT_(2)=(20-11)^(@)C`
= `9^(@)C`
`c_(2)=4.2Jg^(-1)""^(@)C^(-1)`
Heat lost by hot body
= `m_(1)c_(1)DeltaT_(1)`
= `50xxc_(1)xx130` joule
Heat gained by cold body
= `m_(2)c_(2)DeltaT_(2)`
= `100xx4.2xx9` joule
From the principle of calorimetry, if the system if fully insulated, then,
Heat lost by hot body = Heat gained by cold body
or `50xxc_(1)xx130=100xx4.2xx9`
or `c_(1)=(100xx4.2xx9)/(50xx130)Jg^(-1)""^(@)C^(-1)`
= `0.58Jg^(-1)""^(@)C^(-1)`