Question

551 cal of heat is added to 5.00 g ice at –20.0 °c. what is the final temperature of the water?

Answer 1

This item can be answered using the equation,
  
               h = mcp(dT) + mHv

where h is the heat, m is the mass of the substance, cp is the specific heat, and dT is the temperature difference, Hv is the latent heat of fusion. Substituting the known values from the given above,

            551 cal = (5 g)(1 cal/g°C)(T - -20) + (5 g)(80 cal/g)

The value of T from the equation is 10.2.

Answer: 10.2°C