Question

Calculate the amount of heat absorbed by 100 g
of ice to change into water of 30°C. (Specific heat
capacity of water is 1 cal/g°C)
(11000 cal)​

Answer 1

Answer:

Let rise in temperature=θ R

Heatsuppliedtocopper= ss20ca ×120s=2400cal

Also, Heatsupplied=mcθ R

2400cal=0.12kg×100calkg −1 C −1 θ R

∴θ R

= 0.12×1002400⁰C=200⁰ C

The final temperature of the copper =20 C+200o C =220⁰ C.

I hope you got your answer.

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