A) 15%
B) 30%
C) 25%
D) 20%
Correct Answer: D
Solution :
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 \[=\frac{\Delta \theta }{\theta }\times 100=\frac{36.4-30}{30}\times 100=21.33%\]You need to login to perform this action.
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