A mixture consists of 50 g of ice and 60 g of liquid water, both at 0 degree C. The amount of...

Question:

A mixture consists of {eq}50 \ g {/eq} of ice and {eq}60 \ g {/eq} of liquid water, both at {eq}0 ^\circ \ C {/eq}. The amount of heat that must be added to melt all of the ice is about?

Heat Transfer

The heat required for a certain change in the temperature of a substance can be solved using the basic equation given as {eq}Q = mC \Delta T {/eq}, where {eq}m {/eq} is the mass of the substance, {eq}C {/eq} is the specific heat of the substance, and {eq}\Delta T {/eq} is the difference between the final and initial temperature of the material. Additionally, the energy for a change phase is solved using the latent heat equation given as {eq}Q = mL {/eq}, where {eq}L {/eq} is the latent heat of fusion or vaporization of the substance.

We find the energy needed to melt the ice using the equation

{eq}Q = m_{ice \rightarrow water}L_f {/eq}

where

• {eq}L_{f} = 334 \ \rm{J/g} {/eq} the latent heat of fusion of water,
• {eq}m = 50\ g {/eq} is mass of the ice.

Substituting all given values, we get:

{eq}Q = m_{ice \rightarrow water}L_f \\ Q = (50\ g)(334\ J/g) \\ \boxed{Q = 16700\ J} {/eq}