815 cal of heat is added to 5.00 g ice at - 20.0 °C. What is the final temperature of the water?

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Kenya

Chemistry

3 September, 09:32

815 cal of heat is added to 5.00 g ice at - 20.0 °C. What is the final temperature of the water?

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Rowe · Answer 1

The amount of heat needed to raise the temperature of an object is obtained through the equation,

H = m (cp) (20) + m (heat of fusion) + m (cp) (dT)

where H is heat, m is mass, cp is specific gravity, and dT is the change in temperature. The specific gravity of water is 0.5 cal/g. C. The third term is for water and cp is equal to 1 cal/g. C. Substituting the values,

815 cal = (5 g) (0.5 cal/g. C) (20C) + (80 cal/g) (5 g) + 5 (1) (T2)

The value of T2 is 73 degrees C.