InterviewSolution
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InterviewSolution
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`0.1 m^(3)` of water at `80^(@)C` is mixed with `0.3m^(3)` of water at `60^(@)C`. The finial temparature of the mixture isA. `70^(@)C`B. `65^(@)C`C. `60^(@)C`D. `75^(@)C` |
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Answer» Correct Answer - B Density of water = `10^(3)kg m^(-3)` Let the final temperature of the mixture be t. Assuming no heat transfer to or from container. Heat lost by water = `0.1 xx 10^(3) xx S_(water) xx (80-t)` Heat gained by water = `0.3 xx 10^(3) xx S_(water) (t-60)` According to principle of calorimetry Heat lost = heat gain `0.1 xx 10^(3) xx S_(water) xx (80-t) = 0.3 xx 10^(3) xx S_(water) xx (t-60)` `rArr 1 xx(80-t)= 3xx (t-60) rArr t=65^(@)`C |
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