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10 gm of ice at – 20^(@)C is added to 10...

10 gm of ice at `– 20^(@)C` is added to 10 gm of water at `50^(@)C`. Specific heat of water `= 1 cal//g–.^(@)C`, specific heat of ice = `0.5 cal//gm-.^(@)C`. Latent heat of ice = 80 cal/gm. Then resulting temperature is -

A

10 g

B

5 g

C

0 g

D

20 g

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PHYSICS GALAXY - ASHISH ARORA-HEAT AND THERMAL EXPANSION-NUMERICAL MCQs SINGLE OPTIONS CORRECT
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