91Ó°ÊÓ

The given information are used to find the joules:

Mass of steam$=125g$

Initial temperature$={100}^{∘}C$

Cooling temperature$=0^{∘}C$

Steam has a mass of $125g$and a vaporization heat of $2260J/g$As a result, the formula for calculating the energy released to condense $125g$of steam is:

$$\begin{gathered} Heat=mass×heatofvaporization \\ =125g×\frac{2260J}{g} \\ =2.83×{10}^{5}J \end{gathered}$$

Calculate the Heat energy,

$$\begin{gathered} Heat=m×∆t×SH \\ =125g×\left({85}^{∘}C-0^{∘}C\right)×\frac{4.184J}{g^{∘}C} \\ =4.45×{10}^{4}J \end{gathered}$$

The total energy is equal to the sum of the energy is released during condensation and the energy required to chill the liquid. The formula is as follows:

$$\begin{gathered} =2.83×{10}^{5}J+4.45×{10}^{4}J \\ =3.28×{10}^{5}J \end{gathered}$$

Then the Total amount of heat energy in Joule is$3.28×{10}^{5}J$

The following are the information to find Kilocalories:

Mass of Ice$=525g$

Initial temperature$=0^{∘}C$

Heating temperature$={15}^{∘}C$

The heat of fusion of water is $80cal/g$, while the mass of ice is $525g$. As a result, the calculation for calculating the amount of energy required to melt $525g$of ice is as follows:

$$\begin{gathered} Heat=mass×heatoffusion \\ =525g×\frac{80cal}{g} \\ =4.20×{10}^{4}cal \end{gathered}$$

Calculate the Heat energy to warm the liquid from $0^{∘}C$to ${15}^{∘}C$is

$$\begin{gathered} Heat=m×∆t×SH \\ =525g×\left(15.0^{∘}C-0^{∘}C\right)×\frac{1cal}{g^{∘}C} \\ =7.88×{10}^{3}cal \end{gathered}$$

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

$$\begin{gathered} =4.20×{10}^{4}cal+7.88×{10}^{3}cal \\ =5.0×{10}^{4}cal \end{gathered}$$

Therefore, the Total amount of heat in calories is$5.0×{10}^{4}cal$

The following are the information to find Kilojoules is as:

Mass of steam$=85.0g$

Initial temperature$={100}^{∘}C$

Cooling temperature$=0^{∘}C$

Steam's mass is $85g$. the formula to find the Heat released during condensation is as:

$$\begin{gathered} Heat=mass×heatofvaporization \\ =85.0g×\frac{2260J}{g} \\ =1.92×{10}^{5}J \end{gathered}$$

Calculate the Heat energy to cool down the liquid from ${100}^{∘}C$to $0^{∘}C$is as follows:

$$\begin{gathered} Heat=m×∆t×SH \\ =85.0g×\left({100}^{∘}C-0^{∘}C\right)×\frac{4.184J}{g^{∘}C} \\ =3.56×{10}^{4}J \end{gathered}$$

The following is the formula for calculating the amount of energy required to freeze the liquid:

$$\begin{gathered} Heat=mass×heatoffusion \\ =85.0g×\frac{334J}{g} \\ =2.84×{10}^{4}J \end{gathered}$$

The Total Energy calculated as follows:

$$\begin{gathered} 1.92×{10}^{5}J+3.56×{10}^{4}J+2.84×{10}^{4}J=2.56×{10}^{5}J×\frac{1KJ}{1000J} \\ =2.56×{10}^{2}KJ \end{gathered}$$