Question

In their Memoir on Heat,published in 1783, Lavoisier and Laplace reported, “The heat necessary to melt ice is equal to three quarters of the heat that can raise the same mass of water from the temperature of the melting ice to that of boiling water” (English translation). Use this 18th-century observation to compute the amount of heat (in joules) needed to melt 1.00 g ice. Assume that heating 1.00 g

water requires 4.18 J of heat for each$\mathbf{1}\mathbf{.}{\mathbf{00}}^{\mathbf{0}}\mathit{C}$ throughout the range from${\mathbf{0}}^{\mathbf{0}}\mathit{C}$ to ${\mathbf{100}}^{\mathbf{0}}\mathit{C}$.

Short answer

The heat required to melt 1.00 g of ice is 314 J.

Step-by-step solution

The given Information;

Here, two bodies of equal mass and different temperatures are kept in thermal contact. Their specific heat capacities are inversely proportional to the change in temperature undergoing on reaching their final temperature.

The heat

Heat constitutes an important term in thermodynamics denoted as q. There are numerous methods of heat transfer, and they are conduction and convention.

The amount of heat needed to melt 1.00 g of ice.

The amount of heatneeded to raise the 1.00 g of water from the meltingice temperature$\left(0^{0}C\right)$ to that of boiling water is;

$$\begin{gathered} {\text{q = Mc}}_{\text{s}}\text{∆T} \\ \text{= 1.00 g×}\frac{\text{4.18 J}}{{\text{g}}^0\text{C}}\text{×}\left({\text{100}}^0{\text{C - 0}}^0\text{C}\right) \\ \text{= 418 J} \end{gathered}$$

This heat is equal to three-quarters of the heat that can raise the same mass of water by ${100}^{0}C$.

Therefore, theamount of heat neededto melt 1.00 g of ice is:

$$\begin{gathered} \frac{\text{3}}{\text{4}}\text{q =}\frac{\text{3}}{\text{4}}\text{418 J} \\ \text{= 314 J} \end{gathered}$$