# Question #e9134

Jan 17, 2016
1. Temp. of mixture is $= {10.5}^{o} C$
2. Temp. of mixture is $\approx {88.3}^{o} C$

#### Explanation:

As the quantities have been given in CGS units, therefore, answers have been worked out in same units.
Values used
Latent Heat of fusion of water $= 79.5 c a l / g m = 333 k J / k g$
Specific heat of water$= 1 c a l / g m = 4.19 k J / k g$

Latent heat exchanged for change of state ${Q}_{l} = m L$,
where $m \mathmr{and} L$ are mass and latent heat of the substance.

Amount of heat exchanged is given as $\Delta Q = m s \Delta t$,
where $m$ is the mass, $s$ is the specific heat and $\Delta t$ is change in temperature.

Case 1.
Let the final temperature of the mixture be water at ${t}_{1}^{o} C$.
Heat lost by 5gm of water at ${90}^{o} C$ to become Water at ${t}_{1}^{o} C$ is given as

$\Delta {Q}_{l o s t} = 5 \times 1 \times \left(90 - {t}_{1}^{o}\right)$

Simlarly heat gained by ice at ${0}^{o}$ to water at ${t}_{1}^{o} C$ is given as
$\Delta {Q}_{g a i n e d} =$heat gained by $5 g m$ Ice at ${0}^{o} C$ for the change of state$+$Heat gained by 5gm of water at ${0}^{o} C$ to become Water at ${t}_{1}^{o} C$

$\Delta {Q}_{g a i n e d} = 5 \times 79.5 + 5 \times 1 \times \left({t}_{1}^{o} - 0\right)$

Since, $\Delta {Q}_{l o s t} = \Delta H e a {t}_{g a i n e d}$
$5 \times 1 \times \left(90 - {t}_{1}^{o}\right) = 5 \times 79.5 + 5 \times 1 \times \left({t}_{1}^{o} - 0\right)$
$\implies \left(90 - {t}_{1}^{o}\right) = 79.5 + {t}_{1}^{o}$
$\implies 2 {t}_{1}^{o} = {10.5}^{o} C$

Case 2.
Let the final temperature of the mixture be water at ${t}_{2}^{o} C$.
Heat lost by 500gm of water at ${90}^{o} C$ to become Water at ${t}_{2}^{o} C$ is given as

$\Delta {Q}_{l o s t} = 500 \times 1 \times \left(90 - {t}_{2}^{o}\right)$

As in Case 1. heat gained by ice at ${0}^{o}$ to water at ${t}_{2}^{o} C$ is given as
$\Delta {Q}_{g a i n e d} =$heat gained by $5 g m$ Ice at ${0}^{o} C$ for the change of state$+$Heat gained by 5gm of water at ${0}^{o} C$ to become Water at ${t}_{2}^{o} C$

$\Delta {Q}_{g a i n e d} = 5 \times 79.5 + 5 \times 1 \times \left({t}_{2}^{o} - 0\right)$

Since, $\Delta {Q}_{g a i n e d} = \Delta H e a {t}_{l o s t}$
$500 \times 1 \times \left(90 - {t}_{2}^{o}\right) = 5 \times 79.5 + 5 \times 1 \times \left({t}_{2}^{o} - 0\right)$
$\implies 100 \left(90 - {t}_{2}^{o}\right) = 79.5 + {t}_{2}^{o}$
$\implies 101 {t}_{2}^{o} = {8920.5}^{o} C$