91Ó°ÊÓ

The following are the given information as,

Mass of ice $=50.0g$

Initial temperature$=0^{∘}C$

Temperature to warm the liquid$={65}^{∘}C$

The heat of fusion of water is $334J/g$, while the mass of ice is $50.0g$. The following is the formula for calculating the energy required to melt$50.0g$of ice:

$$\begin{gathered} Heatchange=mass×heatoffusion \\ =50.0g×\frac{334J}{g} \\ =16700J \end{gathered}$$

The Heat Equation for the changes from $0^{∘}C$to ${65}^{∘}C$as:

$Heat=m×∆t×SH$

localid="1651762965771" $$\begin{gathered} =50.0g×({65}^{∘}C-0^{∘}C)×\frac{4.184J}{g^{∘}C} \\ =13598J \end{gathered}$$

The total energy is equal to the sum of the energy required to melt and warm the liquid. The formula is as follows:

$16700J+13598J=30298J$

The Amount of Energy will be$30298J$

The given information are as follows:

Mass of steam$=15.0g$

Initial temperature$={100}^{∘}C$

Liquid cools to$0^{∘}C$

The mass of ice is $15.0g$, the heat of fusion of water is localid="1651762439867" $540cal/g$. So, the formula used to calculate the energy needed to $15.0g$melt of ice is as follows:

localid="1651763171187" $$\begin{gathered} Heatchange=mass×heatoffusion \\ =15.0g×\frac{540cal}{g} \\ =8.10×{10}^{3}cal \end{gathered}$$

Heat equation to cool the liquid from ${100}^{∘}C$to $0^{∘}C$as:

$Heat=m×∆t×SH$

$$\begin{gathered} =15.0g×({100}^{∘}C-0^{∘}C)×\frac{1cal}{g^{∘}C} \\ =1.50×{10}^{3}J \end{gathered}$$

The total energy is calculated by adding the energy required to melt and the energy required to warm the liquid. The following is how it's calculated:

$$\begin{gathered} 8.10×{10}^3\mathrm{cal}+1.50×{10}^3\mathrm{cal}=9.60×{10}^3\mathrm{cal}×\frac{0.001\mathrm{kcal}}{1\mathrm{cal}} \\ =9.60Kcal \end{gathered}$$

The Energy will be$9.60Kcal$

The given information are as follows:

Mass of ice$=24.0g$

Initial temperature$=0^{∘}C$

Warm temperature$={100}^{∘}C$

The mass of ice is $24.0g$, the heat of fusion of water is $334J/g$. So, the formula used to calculate the energy needed to melt $24.0g$of ice is as follows:

$$\begin{gathered} Heatchange=mass×heatoffusion \\ =24.0g×\frac{334J}{g} \\ =8.02×{10}^{3}cal \end{gathered}$$

The heat to warm liquid from $0^{∘}C$to ${100}^{∘}C$is,

$Heat=m×∆t×SH$

$$\begin{gathered} =24.0g×({100}^{∘}C-0^{∘}C)×\frac{4.184J}{g^{∘}C} \\ =1.00×{10}^{4}J \end{gathered}$$

The vaporization heat is $2260J/g$. The formula for calculating the amount of energy required to evaporate is,

$$\begin{gathered} Heat=mass×heatofvaporization \\ =24.0g×\frac{2260J}{g} \\ =5.42×{10}^{4}J \end{gathered}$$

The total energy is equal to the sum of the energy required to melt and the energy required to warm the liquid. It's computed like this:

$$\begin{gathered} 8.02×{10}^3\mathrm{J}+1.00×{10}^4\mathrm{J}+5.42×{10}^4\mathrm{J}=7.22×{10}^4\mathrm{J}×\frac{1\mathrm{kJ}}{1000\mathrm{J}} \\ =7.2\mathrm{KJ} \end{gathered}$$

The Heat energy will be$7.22KJ$