# Question f5a12

Mar 22, 2016

=31231J

#### Explanation:

Heat required to raise the temp of ice from $- 19 \to {0}^{o} C$
${H}_{1} = 10.01 \times 2.06 \times \left(0 - \left(- 19\right)\right) \approx 391.79 J$ (Taking Spheat of ice$2.06 \frac{J}{{g}^{o} C}$)
Heat equired to melt ice ice${H}_{2} = 10.01 \times 336 J = 3363.36 J$
(Taking latent heat of fusion of ice$336 \frac{J}{g}$)
Heat required to raise the temp of water from $- {0}^{o} C \to {100}^{o} C$
${H}_{3} = 10.01 \times 4.2 \times 100 = 4204.20 J$

Heat required to vaporize water at its BP
${H}_{4} = 10.01 \times 2268 = 22702.68 J$
(Taking latent heat of vaporization of water$336 \frac{J}{g}$)

Heat required to raise the temp of vapor from ${100}^{o} C \to {128}^{o} C$
${H}_{5} = 10.01 \times 2.03 \times 28 = 568.97 J$
(Taking Spheat of vapor $2.03 \frac{J}{{g}^{o} C}$

TOTAL $H = 31231 J$

Mar 23, 2016

$7442.44 c a l$ rounded to second place of decimal

#### Explanation:

We know that in such thermal interactions 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.
Lets choose CGS units as preferred system.

This a five step process.
1. Heating of ice at $- {19}^{\circ} \text{C}$ to make it ice at ${0}^{\circ} \text{C}$. Taking the specific heat of ice as $0.5 c a l / {g}^{\circ} \text{C}$ and $t = 19$

$\Delta {Q}_{1} = 10.01 \times 0.5 \times 19 = 95.095 c a l$
2. State of change from ice at ${0}^{\circ} \text{C}$ to water at ${0}^{\circ} \text{C}$. Taking latent heat of fusion of ice as $80 c a l / g$

$\Delta {Q}_{2} = 10.01 \times 80 = 800.8 c a l$
3. Heating of water at ${0}^{\circ} \text{C}$ to water at ${100}^{\circ} \text{C}$. Taking the specific heat of water as $1 c a l / {g}^{\circ} \text{C}$ and $t = 100$

$\Delta {Q}_{3} = 10.01 \times 1 \times 100 = 1001 c a l$
4. State of change from water at ${100}^{\circ} \text{C}$ to steam at ${100}^{\circ} \text{C}$. Taking latent heat of evaporation of water as $540 c a l / g$

$\Delta {Q}_{4} = 10.01 \times 540 = 800.8 c a l = 5405.4 c a l$
5. Heating of steam at ${100}^{\circ} \text{C}$ to steam at ${128}^{\circ} \text{C}$. Taking the specific heat of steam as $0.5 c a l / {g}^{\circ} \text{C}$ and $t = 28$

$\Delta {Q}_{5} = 10.01 \times 0.5 \times 28 = 140.14 c a l$

Total heat required is
$\Delta {Q}_{1} + \Delta {Q}_{2} + \Delta {Q}_{3} + \Delta {Q}_{4} + \Delta {Q}_{5} = 7442.435 c a l$

$= 7442.44 c a l$ rounded to second place of decimal