Question

The water of mass 75g at 100 C is added to ice of mass 20g at -15 C. What is resulting temperature approx? (Latent heat of ice = 80Cal/g and specific heat of ice=0.5 cal/g)

Answer 1

Answer:

60⁰C

Explanation:

Let final temperature is T

Then,

Heat Loss=Heat Gain

7500-75T=150+1600+20T

95T=5750

T=60.5⁰C≈60⁰C

Hence this is the required result.

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