Question

Total heat required to convert $1\ g$ of ice at $0^\circ C$ into steam at $100^\circ C$ is $(L_{steam} = 536\ cal/g)$ :
(A) $100\ cal$
(B) $0.01\ kilo\ calorie$
(C) $716\ cal$
(D) $1\ kilo\ calorie$

Answer 1

The conversion of ice into steam takes place in three steps: First the ice converts into the water, then the temperature of the water increases to $100^\circ C$ and then the water evaporates to form steam. The amount of heat required to convert $1\ g$ of ice at $0^\circ C$ into steam at $100^\circ C$ is the sum of heat required to convert ice into water, the heat required to raise the temperature from $0^\circ C$ to $100^\circ C$ and then to convert water at $100^\circ C$ into steam.

Complete step by step answer:
Given, Mass of ice to be converted into steam $= 1\ g$
Temperature of ice $= 0^\circ C$
The heat required to convert ice at $0^\circ C$ into the water at $0^\circ C$ is given as:
Mass of ice $\times$ Latent heat of fusion of ice
Latent heat of fusion of ice $= 80\ cal/g$
Therefore,
Heat required to convert ice at $0^\circ C$ into water at $0^\circ C = 1\ g\times 80\ cal/g = 80\ cal$
Now, the water at $0^\circ C$ has to be first heated to $100^\circ C$ to be changed into steam.
The heat required to raise the temperature of water from $0^\circ C$ to $100^\circ C$ is given as:
Mass of water $\times$ Increase in temperature $\times$ Specific heat of water
Mass of water is the same as the mass of ice, which is equal to $1\ g$.
The increase in temperature of water $= 100^\circ C - 0^\circ C = 100^\circ C$
Specific heat of water $= 1\ cal/g^\circ C$
Putting all the required values, the heat required to raise the temperature of water $= 1\ g\times 100^\circ C\times 1\ cal/g^\circ C = 100\ cal$
After this, further, some heat is required to be supplied to convert the water at $100 ^\circ C$ to steam at the same temperature.
The heat required to convert water into steam $=$ Mass of water $\times$ Latent heat of steam
Mass of water $= 1\ g$
The latent heat of steam is given as $536\ cal/g$.
Therefore, the heat required $=1\ g\times 536\ cal/g = 536 cal$
Now, the total heat required to convert ice to steam $=$ Heat required to convert ice into water $+$ Heat required to raise the temperature to $100^\circ C$ $+$ Heat required to convert water into steam
$=80\ cal + 100\ cal + 536\ cal$
$=716\ cal$
Therefore, the heat required to convert ice at $0^\circ C$ into steam at $100^\circ C$ is equal to $716\ cal$.

So, the correct answer is Option C.

Note: Fusion means melting. So, the latent heat of fusion is defined as the amount of heat required for a solid to melt into a liquid without any temperature rise. Specific heat of any substance is defined as the amount of heat (in calories) required to raise the temperature of one gram of the substance by one degree Celsius.