Question

Given, $19g$ of water at \[30_{}^ \circ C\] and $5g$ of ice at $ - 20_{}^ \circ C$ are mixed together in a calorimeter. What is the final temperature of the mixture? (Given specific heat of ice =$0.5ca\lg _{}^{ - 1}(_{}^ \circ C)_{}^{ - 1}$ and latent heat of fusion of ice =$80ca\lg _{}^{ - 1}$)
A.$0_{}^ \circ C$
B.$ - 5_{}^ \circ C$
C.$5_{}^ \circ C$
D.$10_{}^ \circ C$

Answer 1

Hint: The measuring of heat is known as calorimetry. We will solve this question by first calculating the heat released by water. Then we calculate the heat required to raise the temperature of ice, heat taken by the ice, and heat required to raise the temperature of ice water to the final temperature. And the final step is to use the Principle of Calorimetry.
Formula used:
Specific heat capacity – \[Q = ms\Delta \theta \]
Where $Q$is the heat supplied, $m$ is the mass of substance, $\Delta \theta $ is the temperature change, and $s$ is the specific heat capacity.
Specific latent heat of fusion – \[Q = mL\]
$Q$ is heat need to melt/vaporize, $m$is the mass of the substance, and $L$is the latent heat of fusion.

Complete step by step answer:
The given data in question -
$19g$ of water at \[30_{}^ \circ C\]
$5g$ of ice at $ - 20_{}^ \circ C$
specific heat of ice =$0.5ca\lg _{}^{ - 1}(_{}^ \circ C)_{}^{ - 1}$
Ice's latent heat of fusion =$80ca\lg _{}^{ - 1}$
Using the specific heat formula heat given by water is calculated as
$Q_w^{} = m_w^{}s_w^{}\Delta \theta _w^{}$
$Q_w^{} = (19x1x(30 - \theta _f^{}))$
Where $\theta _f^{}$ is the final temperature of the mixture
$Q_w^{} = 570 - 19\theta _f^{}$
Similarly, we calculate heat taken by ice using specific heat formula, we get,
\[Q_{i(t)}^{} = m_i^{}s_i^{}\Delta \theta _i^{}\]
\[Q_{i(t)}^{} = 5x0.5x(0 - ( - 20))\]
\[Q_{i(t)}^{} = 5x0.5x20\]
\[Q_{i(t)}^{} = 50cal\]
Calculating heat required to melt the Ice by using specific latent heat of fusion, we get,
\[Q_{i(req)}^{} = mL\]
\[Q_{i(req)}^{} = 5x80\]
\[Q_{i(req)}^{} = 400cal\]
Since ice will melt to form cold(icy) water then heat required to raise the temperature of ice water is given by -
\[Q_{iw}^{} = m_i^{}s_w^{}\Delta \theta _{iw}^{}\]
\[Q_{iw}^{} = 5x1x(\theta _f^{} - 0)\]
\[Q_{iw}^{} = 5\theta _f^{}\]
According to the Principle of Calorimeter, heat given by hot substances is equal to the heat received by cold substances.
\[Heat{\text{ }}Lost{\text{ }} = {\text{ }}Heat{\text{ }}Given \\
    \\
 \]\[Heat{\text{ }}Lost = Heat{\text{ }}taken{\text{ }}by{\text{ }}the{\text{ }}ice{\text{ }} + Heat{\text{ }}required{\text{ }}to{\text{ }}melt{\text{ }}ice + Heat{\text{ }}take{\text{ }}by{\text{ }}icy{\text{ }}water\]
\[Q_w^{} = Q_{i(t)}^{} + Q_{i(req)}^{} + Q_{iw}^{}\]
Using the data from the above calculation, we obtain
\[570 - 19\theta _f^{} = 50 + 400 + 5\theta _f^{}\]
\[24\theta _f^{} = 120\]
\[\theta _f^{} = 5_{}^ \circ C\]

Hence the correct option is C) 5.
From the above calculation, we can conclude that heat given by water will melt all the ice as \[Q_w^{} > Q_{i(t)}^{} + Q_{i(req)}^{}\].

Note:
While reading the question please check the units of the given values and change the unit accordingly.
To convert from calorie to joule - $1cal = 4.186J$
The in specific heat, the formula is called heat capacity of the body and its unit is $J/K$.