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JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
- JEE Mains 2013
- Medium
- Numerical
\(500\, g\) of water and \(100\, g\) of ice at \(0\,^oC\) are in a calorimeter whose water equivalent is \(40\, g\). \(10\, g\) of steam at \(100\,^oC\) is added to it. Then water in the calorimeter is ....... \(g\) (Latent heat of ice \(\,= 80\, cal/g\), Latent heat of steam \(\,= 540\, cal/ g\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(590\)
Step-by-step Solution
Detailed explanation
As \(1\, g\) of steam at \(100\,^oC\) melts \(8\,g\) of ice at \(0\,^oC\). \(10\, g\) of steam will melt \(8\times10\, g\) of ice at \(0\,^oC\) Water in calorimeter \(\,= 500 + 80 + 10\,g\, = 590\,g\)
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