Question

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

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.