Question

How much amount of heat is required for ${10^{ - 2}}\,Kg$ of ice at $- {8^0}C$ to convert into steam at ${100^0}C$.
A. $30.30 \times {10^3}\,J$
B. $30.10 \times {10^3}\,J$
C. $30.20 \times {10^3}\,J$
D. $30.40 \times {10^3}\,J$

Answer 1

Here, we have been asked to calculate the amount of heat required to convert a certain amount of ice at a given temperature into steam at its boiling point or at ${100^0}C$. We have to understand that here the concept used is of latent heat of fusion. We have the temperature at which the ice is and then we have to boil it till ${100^0}C$ and make the water to reach the temperature at which it will convert into steam.

Complete step by step answer:
Now according to the given data we have the mass of the ice as ${10^{ - 2}}Kg$ and its temperature ${T_1} = - {8^0}C$. We know the process of converting ice to water and then to steam so we go accordingly. When we bring ice from $- {8^0}C$ to ${0^0}C$ by using the formula for the heat required for this conversion.
$Q = mC\Delta T$ …. $(1)$
Let ${Q_1}$ be the heat for conversion of ice from $- {8^0}C$ to ${0^0}C$, specific heat of ice ${C_{ice}} = 2.06J{\left( {{g^0}C} \right)^{ - 1}}$ and $\Delta T = 0 - ( - 8) = {8^0}C$.
$\Rightarrow {Q_1} = ({10^{ - 2}} \times 2.06 \times 8)J$ …. (in $(1)$ )
$\Rightarrow {Q_1} = 164.8J$ …. $(2)$

Now, to bring ice from ${0^0}C$ to the normal water temperature, we have to use the latent heat of fusion i.e. $\Delta {H_f} = 334J{g^{ - 1}}$ in the formula for heat required for conversion of ice to water and let ${Q_2}$ be that heat.
${Q_2} = m\Delta {H_f}$
Using the above specified values we have:
${Q_2} = \left( {{{10}^{ - 2}} \times 334} \right)J$
$\Rightarrow {Q_2} = 3340J$ …. $(3)$
Now, the heat required to heat water from ${0^0}C$ to ${100^0}C$ and let this heat be ${Q_3}$. So the $\Delta T = {100^0}C$ and use it in equation $(1)$
Specific heat of the water is $4.19J{({g^0}C)^{ - 1}}$
${Q_3} = \left( {{{10}^{ - 2}} \times 4.19 \times 100} \right)J$
$\Rightarrow {Q_3} = 4190J$ …. $(4)$

Now, when water is heated to vaporize or convert into steam by using later heat of vaporization.
$\Delta {H_v} = 2257J{g^{ - 1}}$. Let the heat required be ${Q_4}$
$\Rightarrow {Q_4} = m\Delta {H_v}$
$\Rightarrow {Q_4} = \left( {{{10}^{ - 2}} \times 2257} \right)J$
$\Rightarrow {Q_4} = 22570J$ …. $(5)$
Thus, the total heat required to convert ice to water is given by adding equation $(2)$ , $(3)$ , $(4)$ and $(5)$
$Q = {Q_1} + {Q_2} + {Q_3} + {Q_4}$
$\Rightarrow Q = 164.8J + 3340J + 4190J + 22570J$
$\therefore Q = 30.30 \times {10^3}J$
Thus, the required heat to convert ice from $- {8^0}C$ to steam is $30.30 \times {10^3}J$.

Hence, the correct answer is option A.

Note: The amount of heat most of it depends on the amount of material and the magnitude of heat to be raised to convert material from one state to another. The heat required to raise the temperature of one kilogram of material by one degree kelvin is called the specific heat of that material. When the state of material is changed we use latent heat of fusion.