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2 years ago

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

1 answer:

4 0

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.

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From the question given above, the following data were obtained:

Initial temperature (T₁) = 306 K

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Options B and C

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