# Question #1a163

Apr 25, 2016

$154500 c a l$

#### Explanation:

We know that heat gained/lost is given by
$\Delta Q = m s t$, or $\Delta Q = m L$
where $m , s \mathmr{and} t$ are the mass, specific heat and rise or gain in temperature of the object;
$L$ is the latent heat for the change of state.

In the given problem heat is given to water to increase its temperature from ${22}^{\circ} \text{C}$ to ${100}^{\circ} \text{C}$ and thereafter boil the water to make it steam at ${100}^{\circ} \text{C}$.

1. Water heated from ${22}^{\circ} \text{C}$ to ${100}^{\circ} \text{C}$
Using the expression and taking specific heat of water as 1.
$\Delta Q = m s t$
$\Delta {Q}_{1} = 250 \times 1 \times \left(100 - 22\right) = 19500 c a l$

2. Heat gained by boiling water to change into vapours at the same temperature i.e., ${100}^{\circ} \text{C}$ is given by $\Delta {Q}_{2} = m L$, where Latent heat of vaporization of water is $540 c a l {g}^{-} 1$.
$\therefore \Delta {Q}_{2} = 250 \times 540 = 135000 c a l$

Total heat required $= \Delta {Q}_{1} + \Delta {Q}_{2} = 19500 + 135000$
$= 154500 c a l$