Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The change in energy, 10 grams water, when it gets converted from ${100^0}C$ steam to ${0^0}C$ ice is $\left( {{L_S} = 540{\text{ cal/gm}},{L_i} = 80{\text{ cal/gm}}} \right)$.A. 7.2 KcalB. -1.8 KcalC. -7.2 KcalD. -1.8 Kcal

Last updated date: 13th Jun 2024
Total views: 402.6k
Views today: 6.02k
Verified
402.6k+ views
Hint: In this question, we need to determine the change in the energy when 10 grams of water when it gets converted into ${0^0}C$ ice from ${100^0}C$ steam. For this, we will calculate the heat required for each transition state in the conversion of steam to ice.

Complete step by step answer:${100^0}C$ steam will be converted into ${0^0}C$ ice by following the following transition states.
${100^0}C$ steam first gets converted into ${100^0}C$ of water by following the materialistic change, then, to ${0^0}C$ water by losing the temperature and then to ${0^0}C$ of ice again by following the transition change.
Here, we will calculate the energy (heat) required for each transition.
(I) ${100^0}C$ steam to ${100^0}C$ of water
The product of the mass of the material and the latent heat of enthalpy results in the heat (or energy) required for the transition. Mathematically, $Q = mL$
According to the question, the latent heat of solidification of the steam is ${L_S} = 540{\text{ cal/gm}}$ and $m = 10gm$. So, substitute these values in the formula $Q = mL$ to determine the heat required to convert ${100^0}C$ steam to ${100^0}C$ of water.
${Q_1} = m{L_s} \\ = 10 \times 540 \\ = 5400cal \\$
(II) ${100^0}C$ of water to ${0^0}C$ water
The product of the mass of the material, the latent heat of enthalpy and the change in the temperature results in the heat (or energy) required for the transition. Mathematically, $Q = mL\vartriangle T$
The latent heat of solidification of the water is ${L_w} = 1{\text{ cal/gm}}$ and $m = 10gm$. So, substitute these values in the formula $Q = mL$ to determine the heat required to convert ${100^0}C$ of water to ${0^0}C$ water.
${Q_2} = m{L_w}\vartriangle T \\ = 10 \times 1 \times 100 \\ = 1000cal \\$
(III) ${0^0}C$ water to ${0^0}C$ of ice
The product of the mass of the material and the latent heat of enthalpy results in the heat (or energy) required for the transition. Mathematically, $Q = mL$
According to the question, the latent heat of solidification of the water is ${L_i} = 80{\text{ cal/gm}}$ and $m = 10gm$. So, substitute these values in the formula $Q = mL$ to determine the heat required to convert ${0^0}C$ water to ${0^0}C$ of ice.
${Q_3} = m{L_s} \\ = 10 \times 80 \\ = 800cal \\$
Now, the total heat required is given as the sum of the individual energies.
$Q = {Q_1} + {Q_2} + {Q_3} \\ = 5400 + 1000 + 800 \\ = 7200cal \\ = 7.2Kcal \\$
As the water is changing from steam to ice, it will absorb the energy, so the change in the energy from ${100^0}C$ steam to ${0^0}C$ ice is negative.
Hence, change in the energy required to convert ${100^0}C$ steam to ${0^0}C$ ice is -7.2 Kcal.

Option C is correct.

Note:It is worth noting down here that, the molecular space of the steam is more than that of ice and so, more energy is required to hold the atoms in the ice as compared to steam. So, the energy stored in ice is more and thus, the change in energy is negative.