91Ó°ÊÓ

1. Mass of copper$\left({\mathrm{M}}_{\mathrm{C}}\right)=150\mathrm{g}$
2. Mass of water$\left({\mathrm{M}}_{\mathrm{W}}\right)=220\mathrm{g}$
3. Initial temperature of copper and water is$\left({\mathrm{T}}_{\mathrm{i}}\right)={20}^{∘}\mathrm{C}$
4. Final temperature of the system is$\left({\mathrm{T}}_{\mathrm{f}}\right)={100}^{∘}\mathrm{C}$
5. Mass that is converted to steam is$\left({\mathrm{M}}_{\mathrm{s}}\right)=5.00\mathrm{g}$
6. Latent heat of vaporization is$\left({\mathrm{L}}_{\mathrm{v}}\right)=539\mathrm{cal}/\mathrm{g}$

Specific heat is the amount of heat required to raise the temperature of a unit mass of a substance by one degree. We can find the amount of heat energy required to raise the temperature of water to its boiling point and the amount of heat required to convert some part of the water to steam. Using this heat, we can calculate the temperature of the cylinder as well. We have to use the equation for specific heat and latent heat of vaporization to solve this.

Formula:

The total heat energy required to raise the temperature,$\mathrm{Q}={\mathrm{M}}_{\mathrm{W}}{\mathrm{C}}_{\mathrm{W}}∆\mathrm{T}+{\mathrm{L}}_{\mathrm{V}}{\mathrm{M}}_{\mathrm{S}}$ …(¾±)

Here, Q is heat energy, M_W is mass of water, C_W is specific heat capacity of water,$∆\mathrm{T}$is change in temperature, L_V is latent heat of vaporization, M_S and is mass of steam.

The initial temperature of a heat body,${\mathrm{T}}_{\mathrm{i}}=\frac{{\mathrm{Q}}_{\mathrm{w}}+{\mathrm{Q}}_{\mathrm{b}}}{{\mathrm{C}}_{\mathrm{c}}{\mathrm{M}}_{\mathrm{c}}}+{\mathrm{T}}_{\mathrm{f}}$ …(¾±¾±)

Here, T_i is initial temperature, Q_W is heat required to change the temperature, Q_b heat transferred to the bowl.

The total amount of heat required raising the temperature of water and also to convert the water to vapor can be given using equation (i) and the given values as follows:

$\begin{array}{rcl}{\mathrm{Q}}_{\mathrm{W}}& =& 220\mathrm{g}×\frac{1\mathrm{cal}}{{\mathrm{g}}^{∘}\mathrm{C}}\left[{\left(100-20\right)}^{∘}\mathrm{C}\right]+\left(539\mathrm{cal}/\mathrm{g}×5.00\mathrm{g}\right)\\ & =& 220\mathrm{g}×{80}^{∘}\mathrm{C}+2695\mathrm{cal}\\ & =& 20295\mathrm{cal}\\ & ≈& 20.3\mathrm{kcal}\end{array}$

Hence, the amount of total energy transferred to the environment is 20.3 kcal

Using equation (i) and the given values, the heat transferred to the bowl is given as:

$\begin{array}{rcl}{\mathrm{Q}}_{\mathrm{b}}& =& 0.0923\mathrm{cal}/\mathrm{g}.{}^{∘}\mathrm{C}×150\mathrm{g}×\left({100}^{\mathrm{o}}\mathrm{C}-{20}^{\mathrm{o}}\mathrm{C}\right)\\ & =& 1107.6\mathrm{cal}\\ & ≈& 1.11\mathrm{kcal}\end{array}$

Hence, the total amount of energy transferred to the bowl is 1.11 kcal

Let, the original temperature of the cylinder is${\mathrm{T}}_{\mathrm{i}}$. So, this temperature value can be given using equation (ii) as follows:

$\begin{array}{rcl}{\mathrm{T}}_{\mathrm{i}}& =& \frac{\left(20.3×{10}^3\mathrm{cal}\right)+\left(1.11×{10}^3\mathrm{cal}\right)}{0.0923\mathrm{cal}/{\mathrm{g}}^{\mathrm{o}}\mathrm{C}×300\mathrm{g}}+{100}^{\mathrm{o}}\mathrm{C}\\ & =& \frac{21.41}{27.69}+100\\ & =& {873}^{\mathrm{o}}\mathrm{C}\end{array}$

Hence, the original temperature is${873}^{\mathrm{o}}\mathrm{C}$.