6.2 Kg of metal at 100°C is cooled by 1Kg of water at 0°C . If specific heat capacity of metal is 1/2 of specific heat capacity of water, final temperature of mixture would be

1)50°C

2)68°C

3)70°C

4)80°C

Asked by jhajuhi19 | 16th Apr, 2019, 02:52: PM

Expert Answer:

Let us assume the final temperature to be T
Specific heat capacity  of water is s and that of metal is s
Heat provided by water = ms(T-0 ) = msT = 1×s×T = sT
Heat received by metal = ms(100-T)/2 = 6.2×s×(100-T)/2 = 3.1×s×(100-T)
In this question heat provided by water is equal heat recipe by metal
sT = 3.1×s×(100-T)
Solving this equation we get T = 75.6 Celcius 

Answered by Ankit K | 17th Apr, 2019, 11:02: PM

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