Question

150 g of ice is mixed with 100 g of water at temperature $80{}^\circ$. The latent heat of ice is 80 cal/g and the specific heat of water is $1\text{ }cal/g{}^\circ C$. Assuming no heat lost to the environment, the amount of ice that does not melt is
A. 100 g
B. 0 g
C. 150 g
D. 50 g

Answer 1

Hint: We need to use the main principle of calorimetry. We are to find the heat gained by the cooler body and the heat lost by the hotter body. Then equaling them, a solution is found.

Formula used:
$H_s=mst$
$H_L=mL$

Complete step by step solution:

Let,
$m_w$= mass of water = 100g,
s = specific heat of water = $1\text{ }cal/g{}^\circ C$,
L= latent heat of ice = 80 cal/g.

Also let ‘x’ g of ice be melted.

Now, let $t{}^\circ C$ be the final temperature at equilibrium condition and let ‘x’ g of ice melt. Now, heat rejected by water $H_1=m_ws(80-t)=100(80-t)$. And heat used by ice (in melting + getting heated up to $t{}^\circ C$),

$H_2=xL+xs(t-0)=x(80+t)$

Now, from principle of calorimetry, $H_1=H_2$

$\Rightarrow x\left( 80+t \right)=100\left( 80-t \right)$

Notice that, in the question nothing is mentioned about the final temperature $t{}^\circ C$. So it’s best to take $t{}^\circ C=0{}^\circ C$, since some ice is present even after the process. So, we get,

$x\text{ }=100g$
Remaining ice $=\left( 150-x \right)g=50\text{ }g$

Hence, Option D is the correct answer.

Additional information:
The latent heat of melting of ice is equal to the latent heat of freezing of water. The difference is that heat is required while melting and rejected while freezing.

Note: While considering problems with ice, know that it has to turn into water first. In this part, latent heat is considered. Later, its temperature starts to increase where specific heat is considered. Also keep in mind that while latent heat is exchanged, the temperature does not change.