| A metal ball of mass 0.1 kg is heated upto \[500{}^\circ C\] and dropped into a vessel of heat capacity\[800\,\,J{{K}^{-1}}\] and containing 0.5 kg water. The initial temperature of water and vessel is \[30{}^\circ C\]. What is the approximate percentage increment in the temperature of the water? [JEE Main 11-Jan-2019 Evening] |
| [Specific heat capacities of water and metal are, respectively, |
| 4200 \[Jk{{g}^{-1}}\,{{K}^{-1}}\]and 400 \[J\,k{{g}^{-1}}\,{{K}^{-1}}\]] |
A) 15%
B) 30%
C) 25%
D) 20%
Correct Answer: D
Solution :
| [d] Heat lost by metal ball = Heat gained by container and water |
| \[{{m}_{b}}{{s}_{b}}(500-\theta )={{m}_{c}}{{s}_{c}}(\theta -30)+{{m}_{w}}{{s}_{w}}(\theta -30)\] |
| \[0.1\times 400(500-\theta )=800(\theta -30)+0.5\times 4200(\theta -30)\]\[40(500-\theta )=2900(\theta -30)\] |
| \[20000-40\theta =2900\theta -87000\] |
| \[\theta =\frac{107000}{2940}={{36.4}^{o}}C\] |
| % increase in the temperature of water |
| \[=\frac{\Delta \theta}{\theta }\times 100=\frac{36.4-30}{30}\times 100=21.33%\] |
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