Question

2 kg ice at -10 degree Celsius has to be converted into steam of 100 degree Celsius, calculate the heat required.

Answer 1

In this question we have been asked to calculate the amount of heat is required to convert the ice t -10 degree Celsius to steam of 100 degree Celsius. Now, we know that the ice shall be first brought to zero degree Celsius. Then using the latent heat of fusion converted into water. This water shall then use heat to heat itself up to 100 degree Celsius and at last, the water at this temperature, will be converted into steam using the latent heat. Therefore, the total heat required for the process will be the sum of heat required in each individual process.

Formula Used:
$Q=mC\Delta T$
$Q=m\Delta H$

Complete step by step answer:
Now, we know the process to convert ice into steam at 100 degree Celsius.
Therefore,
Let ${{Q}_{1}}$ be the heat to bring ice from -10 degree Celsius to 0 degree Celsius.
Now, from specific heat formula we know,
$Q=mC\Delta T$
We know that specific heat of ice ${{C}_{ice}}$is $2.06\text{ }J/g{}^\circ C$, the temperature of ice to be brought down from $-10{}^\circ$to $0{}^\circ$. Therefore, $\Delta T=(0-(-10))$
After substituting the values
We get,
${{Q}_{1}}=2\times 2.06\times 10$
$\Rightarrow {{Q}_{1}}=41.2J$ ………….. (1)
Now, the ice at $0{}^\circ$C will be converted into water by using the latent heat of fusion. Let ${{Q}_{2}}$ be the heat required for fusion.
Therefore, from the formula of latent heat of fusion we know,
${{Q}_{2}}=m\Delta {{H}_{f}}$
We know that latent heat of fusion of water is $\Delta {{H}_{f}}~=\text{ }334\text{ }J/g$
Therefore,
${{Q}_{2}}=2\times 334$
$\Rightarrow {{Q}_{2}}=668J$ ……………………….. (2)
Now, the heat required to heat water from $0{}^\circ$ to 100 degree Celsius be${{Q}_{3}}$. Now, we know the $\Delta T=(100-0)$
Therefore,
${{Q}_{3}}=mC(100-0)$
We know that specific heat of water is $4.19\text{ }J/g{}^\circ C$
Therefore, we get,
$\Rightarrow {{Q}_{3}}=2\times 4.19\times 100$
$\Rightarrow {{Q}_{3}}=838J$ ……………… (3)
Now, the water at 100 degree Celsius will be converted to steam using the latent heat of evaporation. We know that latent heat of evaporation of water is given by,
$\Delta {{H}_{v}}~=\text{ }2257\text{ }J/g$
Now, from the formula of latent heat
$\Rightarrow {{Q}_{4}}=m\Delta {{H}_{v}}$ ………………… (${{Q}_{4}}$ is the required for conversion of water to steam)
Therefore,
${{Q}_{4}}=2\times 2257$
Therefore,
$\Rightarrow {{Q}_{4}}=4514J$ …………………. (4)
Now, we know that the total heat required to convert 2 kg ice from -10 degree Celsius to 100 degree Celsius is the sum of heat required by the individual process.
Therefore,
$Q={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}+{{Q}_{4}}$
From (1), (2), (3) and (4) we get,
$\Rightarrow Q=41.2+668+838+4514$
$\Rightarrow Q=6061.2J$

Therefore, the total energy required is 6061.2 Joules.

Note:
The amount of heat required to raise the temperature of a material depends on the mass of the material and the magnitude of temperature change. Therefore, the heat required to raise a temperature of 1 kg of water to 1 degree kelvin is known as specific heat of water. When the state of the material is changed i.e. from solid to liquid we use latent heat of fusion. Similarly, when the state is changed from liquid to gaseous we use latent heat of evaporation. When using the latent heat the temperature does not change.