Calculating the energy in joules required to heat a 256 grames of water at -107 degrees to 333...

Question:

Calculating the energy in joules required to heat a 256 grames of water at -{eq}107^{\circ} {/eq} to {eq}333^{\circ}C {/eq}.

Latent Heat

Latent heat is heat absorbed or released by a substance during a process where the temperature doesn't change. It is associated with phase changes. An example is melting. Heating a solid increases its temperature up to its melting point, but during the process of melting, the temperature does not change, yet heat is absorbed.

This problem requires 5 calculations and each has either a heat capacity or a heat of phase transition associated with it:

1.The heat required to raise the water from -107 to 0 deg. C, using the heat capacity of ice (2.09J/(g deg. C).

2.The heat required to melt the ice using the latent heat of fusion of ice/water (334 J/g)

3.The heat required to heat the water to its boiling point of 100 deg C. The heat capacity of liquid water is 4.186 J/(g deg. C).

4.The heat required to vaporize the water. The latent heat of vaporization of water is 2260 J/g)

5.The heat required to raise the temperature of the water vapor to 333 deg. C. The heat capacity of water vapor is 1.996 J/(g deg. C).

For the entire process, we will be adding heat to water, so all of the heats associated will be positive. We simply need to evaluate each one an add them together. In each case we have the same 256 grams:

1.

{eq}256\;g\times \dfrac{2.09\;J}{g^{\circ}C}\times 107^{\circ}C=57,249.3\;J {/eq}

2.

{eq}256\;g\times \dfrac{334\;J}{g}=85,504J {/eq}

3.

{eq}256\;g\times \dfrac{4.186\;J}{g^{\circ}C}\times 100^{\circ}C=107,161.6\;J {/eq}

4.

{eq}256\;g\times \dfrac{2260\;J}{g}=578,560\;J {/eq}

5.

{eq}256\;g\times \dfrac{1.996\;J}{g^{\circ}C}\times 233^{\circ}C=119,057.4\;J {/eq}

The total heat is the sum of the heats of each process:

The heat required to raise 256 g of water from =107 to 333 deg. C is 947,532.3 J or ~947.5 kJ