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A metal ball of mass 0.1 Kg is heated ...

A metal ball of mass 0.1 Kg is heated upto `500^(@)C` and drooped into a vessel of heat capacity `800 JK^(-1)` and containing 0.5 kg wate. The initial temperature of water and vessel is `30^(@)C`. What is the approximate percentage increment in the temperature of the water ? [Specific Heat Capacities of wate and metal are, repectively,`4200 Jkg^(-1) and 400 Jkg^(-1) K^(-1)`

A

`30%`

B

`25%`

C

`15%`

D

`20%`

Text Solution

Verified by Experts

The correct Answer is:
D
Considering the subscript for ball as .b., for water as .w. and for container as .d. and applying principle of calorimetry (assuming final temperature = `T^@C`)
`m_bs_b(500-T)=m_w s_w(T-30)+m_c s_c (T-30)`
`therefore 0.1 xx 400 (500-T)=0.5 xx 4200(T-30)+800(T-30)`
`therefore 20000-40T=2100T-63000+800T-24000`
`therefore 2940 T=107000`
`therefore T=(10700)/(294)=36.4^@C`
% rise in temperature =`(6.4)/(30)xx100% =21%`
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