# How much heat would be reqired to melt 10.0 g of ice at 0 oC, warm the resulting liquid to 100 oC, and change it to steam at 110 oC?

Mar 17, 2018

$7217$ calories

#### Explanation:

We know latent heat of melting of ice is $80$ calories/$g$

So,to convert $10 g$ of ice at ${0}^{\circ} C$ to same amount of water at the same temperature, heat energy required would be $80 \cdot 10 = 800$ calories.

now,to take this water at ${0}^{\circ} C$ to ${100}^{\circ} C$ heat energy required will be $10 \cdot 1 \cdot \left(100 - 0\right) = 1000$ calories (using,$H = m s d \theta$ where,$m$ is the mass of water,$s$ is specific heat,for water it is $1$ C.G.S unit,and $d \theta$ is the change in temperature)

Now,we know,latent heat of vaporization of water is $537$ calories/$g$

So,to convert water at ${100}^{\circ} C$ to steam at ${100}^{\circ} C$ heat energy required will be $537 \cdot 10 = 5370$ calories.

Now,to convert steam at ${100}^{\circ} C$ to ${110}^{\circ} C$,heat energy required will be $10 \cdot 0.47 \cdot \left(110 - 100\right) = 47$ calories (specific heat for steam is $0.47$ C.G.S units)

So,for this entire process heat energy required will be $\left(800 + 1000 + 5370 + 47\right) = 7217$ calories