Heat energy of 184 kJ is given to ice of mass 600 g at . Specific heat of ice is 2222.3 and latent heat of ice is . (A) Final temperature of the system will be [2023]

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Solution

Q.
Heat energy of 184 kJ is given to ice of mass 600 g at $- 12 ° C$ . Specific heat of ice is 2222.3 $J$ ${kg}^{- 1^{o}}$ $C^{- 1}$ and latent heat of ice is $336$ $kJ / {kg}^{- 1}$ . [2023]

(A) Final temperature of the system will be $0 ° C$ .

(B) Final temperature of the system will be greater than $0 ° C$ .

(C) The final system will have a mixture of ice and water in the ratio of 5 : 1.

(D) The final system will have a mixture of ice and water in the ratio of 1 : 5.

(E) The final system will have water only.

Choose the correct answer from the options given below:

1 A and D only

2 B and D only

3 A and E only

4 A and C only

Ans.
(1)

$Δ Q = 184 \times 10^{3}$

$m = 0.600 kg at - 12 ° C$

$S = 2222.3 J/kg/ ° C$

$L = 336 \times 10^{3} J/kg$

$Q_{1} = 0.600 \times 2222.3 \times 12 = 16000.56 J$

Remaining heat:

$Δ Q_{1} = 184000 - 16000.56 = 167999.44 J$

For melting at $0 ° C$

$Δ Q_{2} = 0.600 \times 336000 = 201600 J (needed)$

$∴$ $100% ice is not melted$

Amount of ice melted

$167999.44 = m \times 336000$ $= 0.4999 kg$

$∴$ $Mass of water = 0.4999 kg$

$Mass of ice = 0.1001 kg$

$∴$ $Ratio = \frac{0.1001}{0.4999} \approx 1 : 5$

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