# Given the Following: heat of fusion of ice = 80.0 cal/g; specific heat of ice = 0.480 cal/g; and specific heat of water = 1.00 cal/g. Calculate the amount of heat required to melt 30.0 g of ice at 0 degrees celsius?

Jul 4, 2017

We wish to assess the enthalpy change for the physical reaction......

#### Explanation:

${H}_{2} O \left(s\right) + \Delta \rightarrow {H}_{2} O \left(l\right)$.......

Note that BOTH the ice and water have a specified temperature of $0$ ""^@C, and thus we assess the heat involved in the phase change, and thus we only need the $\text{latent heat of fusion of ice}$, $80.0 \cdot c a l \cdot {g}^{-} 1$.

And for the given quantity, this $80 \cdot c a l \cdot {g}^{-} 1 \times 30 \cdot g = 2400 \cdot c a l$; whatever a $\text{calorie}$ is...........

By the same token, we would report the enthalpy change of the reverse reaction......

${H}_{2} O \left(l\right) + \Delta \rightarrow {H}_{2} O \left(s\right)$, $\Delta {H}^{\circ} = - 2400 \cdot c a l$. Do you agree?