Question

How much heat is added to 10.0 g of ice at $-200{}^\circ C$to convert it into steam at $120.00{}^\circ C$?

Answer 1

The conversion of ice into steam takes place in phases between solid, liquid and gaseous. The substance in transition state possesses some specific heat in each phase, due to change in energies.
Formula used:
$Q=mc\Delta T$

Complete answer:
Ice is converted into steam by heating it, during which it turns from solid to liquid, then liquid to gas, then finally from gas to steam. These changes take place at different temperatures. At temperature from $-200{}^\circ C$to $0{}^\circ C$, solid ice starts to heat. At $0{}^\circ C$, it starts converting to liquid. Between $0{}^\circ C$ to $100{}^\circ C$ it is liquid and changes to gas after $100{}^\circ C$. Then between $100{}^\circ C$to $120{}^\circ C$, it turns into steam.
Now to calculate the total heat used in this process, we have to add up the heat energy at each phase. For obtaining heat, the expression, $Q=mc\Delta T$is used, where Q is heat energy, m is the mass of substance, c is the specific heat capacity, and $\Delta T$is the change in temperature at each phase.
The heat of various phases is known for pure water as:
Solid specific heat=$2.05\,J/g{}^\circ C$
Liquid specific heat=$4.178\,J/g{}^\circ C$
Vapor specific heat= =$1.89\,J/g{}^\circ C$
Heat of fusion==$333.55\,J/g$
Heat of vaporization=$2257\,J/g$
The heat energies phases along with latent heat of fusion and vaporization are added up to obtain the total heat.
So, total heat Q= ${{(mc\Delta T)}_{solid}}+m{{Q}_{fusion}}+{{(mc\Delta T)}_{liquid}}+m{{Q}_{vaporization}}+{{(mc\Delta T)}_{vapor}}$
Putting the values of all the heat and specific heats in the above formula we get

& Q=10.0g\times 2.05J/g{}^\circ C\times [0-(-200)]{}^\circ C+ \\\ & 10.0g\times 333.55J/g+10.0g\times 4.178J/g{}^\circ C\times [100-0]{}^\circ C+ \\\ & 10.0g\times 2257J/g+10.0g\times 1.89J/g{}^\circ C\times [120-100]{}^\circ C \\\ \end{aligned}$$ $$\begin{aligned} & Q=10.0g\times (2.05J/g{}^\circ C\times 200{}^\circ C)+333.55J/g+ \\\ & (4.178J/g{}^\circ C\times 100{}^\circ C)+2257J/g+(1.89J/g{}^\circ C\times 20{}^\circ C) \\\ \end{aligned}$$ $$Q=10.0g\times (410J/g+333.55J/g+417.8J/g+2257J/g+37.8J/g)$$ Q=34561.5 Joules Hence, the total heat of all the transitions is 34561.5 Joules. **Note:** As the phase changes from solid to liquid, then heat of fusion is also added for it. Similarly, when the state changes from liquid to gas, then heat of vaporization is added. So these heat values should be noted and added for total heat.