# How much heat is required to convert 15.0 g of ice at -15.0 degree C to steam at 100.0 degree C?

## Question:

How much heat is required to convert {eq}15.0 \ g {/eq} of ice at {eq}-15.0 ^\circ \ C {/eq} to steam at {eq}100.0 ^\circ \ C {/eq}?

## Heat Transfer:

When the temperature of an object changes, then the below shows mathematical relation is used to calculate the heat transfer.

{eq}Q=mC\left( {{T}_{1}}-{{T}_{2}} \right) {/eq}; Where,

• Q is the total heat transfer from the object
• m is the total mass
• C is the specific heat capacity
• {eq}{T_1} {/eq} is the initial temperature
• {eq}{T_2} {/eq} is the final temperature

Given data

• The total mass of the ice is {eq}m=15\ \text{g} {/eq}
• The initial temperature is {eq}{{T}_{i}}=-15\ \text{degree}\ \text{Celsius} {/eq}
• The final temperature is {eq}{{T}_{f}}=100\ \text{degree}\ \text{Celsius} {/eq}

The mathematical expression which is generally used to calculate the total heat required to convert ice to steam.

{eq}\begin{align*} Q &={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}+{{Q}_{4}} \\ &=m{{C}_{i}}\left( {{T}_{0}}-{{T}_{i}} \right)+m{{L}_{f}}+m{{C}_{w}}\left( {{T}_{f}}-{{T}_{0}} \right)+m{{L}_{v}} \end{align*} {/eq}

Here, {eq}{Q_1} {/eq} is the heat required to raise the temperature of ice from -12.5 degree Celsius to 0 degree Celsius, {eq}{Q_2} {/eq} is the heat required to convert the ice from 0 degree Celsius to water 0 degree Celsius, {eq}{Q_3} {/eq} is the heat required to raise the temperature of water from 0 degree Celsius to 100 degrees Celsius and {eq}{Q_4} {/eq} is the heat required to convert the water to steam.

Note- The specific heat of ice is {eq}{{C}_{i}}=2.09\ \text{Joule/gram}\cdot \text{degree}\ \text{Celsius} {/eq}, the specific heat of water is {eq}{{C}_{w}}=4.18\ \text{Joule/gram}\cdot \text{degree}\ \text{Celsius} {/eq}, the specific heat of steam is {eq}{{C}_{s}}=2.09\ \text{Joule/gram}\cdot \text{degree}\ \text{Celsius} {/eq}, the fusion heat of water is {eq}{{L}_{f}}=334\ \text{J/g} {/eq}, the vaporization heat of water is {eq}{{L}_{v}}=2257\ \text{J/g} {/eq} and the temperature of water is {eq}{{T}_{0}}=0\text{degree}\ \text{Celsius} {/eq}.

Substitute all the values in the above equation and calculate the total heat required as,

{eq}\begin{align*} Q &=m{{C}_{i}}\left( {{T}_{0}}-{{T}_{i}} \right)+m{{L}_{f}}+m{{C}_{w}}\left( {{T}_{f}}-{{T}_{0}} \right)+m{{L}_{v}} \\ &=\left( 15 \right)\left( 2.09 \right)\left( 0+15 \right)+\left( 15 \right)\left( 334 \right)+\left( 15 \right)\left( 4.18 \right)\left( 100-0 \right)+\left( 15 \right)\left( 2257 \right) \\ &=45605.25\ \text{J} \\ \end{align*} {/eq}

Thus, the total heat required to convert ice to steam is 45605.25 Joules.