Question

The amount of heat required to convert ${\text{1g}}$ of ice $\left( {{\text{Specific}}\,\,{\text{0}}{\text{.5cal}}\,\,{\text{at}}\,\,{{\text{g}}^{{\text{ - 1}}}}{}^{\text{o}}{{\text{C}}^{{\text{ - 1}}}}} \right)$ at ${\text{ - 1}}{{\text{0}}^{\text{o}}}{\text{C}}$ to steam at ${\text{10}}{{\text{0}}^{\text{o}}}{\text{C}}$ is ……………
Given: Latent heat of ice is ${\text{80Cal/gm}}$ , Latent heat of steam is ${\text{540Cal/gm}}$ , Specific heat of water is ${\text{1Cal/gm/C}}$ .
$\left( {\text{a}} \right){\text{ 725cal}}$
$\left( {\text{b}} \right){\text{ 636cal}}$
$\left( {\text{c}} \right){\text{ 716cal}}$
$\left( {\text{d}} \right){\text{ None of these}}$

Answer 1

To answer this question one should have the knowledge of a calorimeter. Calorimeter is the branch of science which deals with the changes in the heat energy of a body. Greater the temperature of the body, more is the heat energy of the body.

Complete Step by step solution:
Given that, latent heat of ice is ${\text{80cal/gm}}$
Latent heat of steam to steam is ${\text{540cal/gm}}$
Specific heat of water is ${\text{1cal/gm/C}}$
We just have to find the amount of heat required to convert ${\text{1g}}$ of ice to steam (vapor).
As, we already know that ${{\text{0}}^{\text{o}}}{\text{C}}$is the point where increase in the temperature will change the state of ice from solid to liquid and with the decrease in temperature form ${{\text{0}}^{\text{o}}}{\text{C}}$ formation of ice will take place.
And, from ${{\text{0}}^{\text{o}}}{\text{C}}$ to ${\text{10}}{{\text{0}}^{\text{o}}}{\text{C}}$ there will be no change in the state there will be only change in the temperature of the water which is formed from the melting of ice at ${{\text{0}}^{\text{o}}}{\text{C}}$ with the increase in temperature. Form ${\text{10}}{{\text{0}}^{\text{o}}}{\text{C}}$ onwards there will be a change in a state will take place in which water is changing into vapor (liquid state to gaseous state).
Now, the calculation is easy.
Amount of heat required is equal to the change of temperature of ice from ${\text{ - 1}}{{\text{0}}^{\text{o}}}{\text{C}}$ to ${{\text{0}}^{\text{o}}}{\text{C}}$ plus the heat required to melt the ice and also the addition of heat required to increase the temperature of water from ${{\text{0}}^{\text{o}}}{\text{C}}$ to ${\text{10}}{{\text{0}}^{\text{o}}}{\text{C}}$ plus the heat required to convert water ${\text{10}}{{\text{0}}^{\text{o}}}{\text{C}}$ to vapor at ${\text{10}}{{\text{0}}^{\text{o}}}{\text{C}}$
Amount of heat required $= 1 \times 0.5[0 - ( - 10)] + 1 \times 80 + 1 \times 1 \times 100 + 1 \times 540$
Amount of heat required $= 5 + 80 + 100 + 540$
Amount of heat required $= 725cal$

Therefore, option $\left( {\text{a}} \right)$ is the correct option.

Note: Learn the basic properties which include physical and chemical properties of ice which will help is future problem solving. Like with the increase and decrease at ${{\text{0}}^{\text{o}}}{\text{C}}$ will result in formation of water and ice respectively and also increase at ${\text{10}}{{\text{0}}^{\text{o}}}{\text{C}}$ will result in formation of water.