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50 g of copper is heated to increase its temperature by `10^@C`. If the same quantity of heat is given to `10g` of water, the rise in its temperature is (specific heat of copper`=420J//kg^(@)//C`)A. `5^(@)C`B. `6^(@)C`C. `7^(@)C`D. `8^(@)C` |
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Answer» Correct Answer - A Same amount of heat is supplied to copper and water. So, `m_(C)c_(C)DeltaT_(c)= m_(W)c_(W)DeltaT` `rArr` `DeltaT_(W) = (m_(C)c_(C)(DeltaT)_(c))/(m_(W)c_(W))` =`(50 xx 10^(-3) xx 420 xx 10)/(10 xx 10^(-3) xx 4200)`= 5^(@)C` |
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