Question

How much heat must be absorbed by ice of mass m=720 at -10°c to take to the liquid state at 15°c​

Answer 1

Q = ms delta theta for ice + ml + msdelta theta for water

Q = 720( 0.5*10 + 80 + 1*15)calories

Q= 720(100)/1000 calories

(720 was in grams converted to kg)

Q = 72 calories or approx 288 J

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