2 kg ice at -20° C is mixed with 5 kg water at 20° C . Then final amount of water in the mixture would be; Given specific heat of ice =0.5 cal/g° C , specific heat of water =1 cal/g° C, Latent heat of fusion of ice =80 cal/g. (in kg )

Question

2 kg ice at -20° C is mixed with 5 kg water at 20° C . Then final amount of water in the mixture would be; Given specific heat of ice =0.5 cal/g° C , specific heat of water =1 cal/g° C, Latent heat of fusion of ice =80 cal/g. (in kg ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

In this problem, ice will need energy to achieve the temperature of

0^{\circ} C

and water will release energy to achieve the temperature of

0^{\circ} C

.
Now, heat lost by

5 k g

water when its temperature decreases from

2 0^{\circ} C

to

0^{\circ} C

is given by

Q_{1} = m s Δ T = (5 \times (10)^{3}) (1) (20)

= 1 0^{5} c a l

And, heat required by

2 k g

ice when its temperature increases from

- 2 0^{\circ} C

to

0^{\circ} C

Q_{2} = m s Δ T = (2 \times (10)^{3}) (0.5) (20)

= 0.20 \times 1 0^{5} c a l

Now, the remaining energy

= Q = Q_{1} - Q_{2} = 0.8 \times 1 0^{5} c a l

will melt some ice. Mass of ice melted is given by using the expression

Q = m L

\Rightarrow m = \frac{Q}{L} = \frac{0.8 \times 1 0 ^{5}}{80} = 1000 g = 1 k g

Hence, the equilibrium temperature of the mixture will be

0^{\circ} C

and mass of the water in equilibrium is is

5 + 1 = 6 k g

.

Q. 2kg ice at −20∘C is mixed with 5kg water at 20∘C . Then final amount of water in the mixture would be; Given specific heat of ice =0.5cal/g∘C , specific heat of water =1cal/g∘C, Latent heat of fusion of ice =80cal/g. (in kg )