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# How much amount of heat is required to convert $1Kg$ of ice at $- 10^\circ C$ into steam $100^\circ C$ at normal pressure?

Last updated date: 20th Sep 2024
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Hint:When the ice is converted into steam it requires heat and it goes through many stages. The specific heat capacity is the concept that is used here. The specific heat capacity is defined as the average amount of heat required to increase the temperature of the $1Kg$ substance by one kelvin. The one kelvin is equal to $- 273^\circ C$. The other concept is called the latent heat of fusion. The latent heat of fusion is the amount of heat required to change the substance from one phase to another phase.

Formula used:
1. $\Delta Q = cm\Delta T$
2. $\Delta Q = mL$
The first formula is used when the heat is required for changing in a single particular phase and the second formula is used while the substance is transformed from one phase to another phase.
Where, $\Delta Q$= Amount of heat required, C= specific heat capacity (For ice,$c = 2100Jk{g^{ - 1}}{k^{ - 1}}$.For water, $c = 4200Jk{g^{ - 1}}{k^{ - 1}}$)
$\Delta T$= Temperature difference, $m$= mass of the substance and $L$= latent heat of fusion (For ice, $L = 3.36 \times {10^5}J{k^{ - 1}}$.For water, $L = 2.25 \times {10^6}J{k^{ - 1}}$)

Here we can divide the phase of the ice while it is converting from ice into steam in four stages. First convert the ice at $- 10^\circ C$ to the ice at $0^\circ C$. In this process, the temperature is changed for the phase from the substance at some phase. Therefore the specific heat capacity concept can be applied here. Therefore the heat changing can be found by,
$\Delta Q = cm\Delta T$
The temperature difference, $\Delta T = {T_2} - {T_1}$
$\Rightarrow \Delta T = 0^\circ C - ( - 10^\circ C)$
$\Delta T = 10^\circ C$
Specific heat capacity of ice, $c = 2100Jk{g^{ - 1}}{k^{ - 1}}$
Mass of the substance, m=$1kg$
Applying these values in the formula gives,
$\Rightarrow \Delta Q = 2100 \times 1 \times 10^\circ C$
$\therefore \Delta {Q_1} = 21000J$ ---------- (1)
Now in stage two, we have to convert the ice at $0^\circ C$ to the water at $0^\circ C$.Here the formula used is $\Delta Q = mL$. The Mass of the substance is not changed. The latent heat of ice$L = 3.36 \times {10^5}J{k^{ - 1}}$
$\Rightarrow \Delta {Q_2} = 1 \times 3.36 \times {10^5}$
$\therefore \Delta {Q_2} = 3.36 \times {10^5}J$ ------------ (2)

In stage three, we have to convert the water at $0^\circ C$to the water at $100^\circ C$. Therefore the formula used is $\Delta Q = cm\Delta T$. The temperature difference, $\Delta T$=$100^\circ C$and the specific heat capacity of water$c = 4200Jk{g^{ - 1}}{k^{ - 1}}$.
$\Rightarrow \Delta {Q_3} = 4200 \times 1 \times 100$
$\therefore \Delta {Q_3} = 4.2 \times {10^5}J$ ------------- (3)
In stage four, we convert the water at $100^\circ C$to the steam at $100^\circ C$ . Therefore the formula used is $\Delta Q = mL$
$\Rightarrow \Delta {Q_4} = 1 \times 2.25 \times {10^6}$
$\therefore \Delta {Q_4} = 2.25 \times {10^6}J$ ---------- (4)
we can find the total heat required by summing the heat required in the four stages. Therefore the total heat required,
$\Delta Q = \Delta {Q_1} + \Delta {Q_2} + \Delta {Q_3} + \Delta {Q_4}$
$\Rightarrow \Delta Q = 21000J + 3.36 \times {10^5}J + 4.2 \times {10^5}J + 2.25 \times {10^6}J$
$\therefore \Delta Q = 3.027 \times {10^6}J$

Hence the total amount of heat required for the $1kg$ ice at $- {10^\circ }C$ to the steam at $100^\circ C$ is $3.027 \times {10^6}J$.

Note:When the water at one phase can be changed into another phase can be done by raising or lowering the temperature of the water at a particular phase. The average amount of heat required for increasing the temperature of one kilogram of a substance by one kelvin is called specific heat capacity. And the heat required to change the phase of the substance is called latent heat.