10-1.Thermometry, Thermal Expansion and Calorimetry
hard

Heat energy of $184\,kJ$ is given to ice of mass $600\,g$ at $-12^{\circ}\,C$, Specific heat of ice is $2222.3\,J\,kg ^{-1^{\circ}}\,C ^{-1}$ and latent heat of ice in $336\,kJ / kg ^{-1}$

$(A)$ Final temperature of system will be $0^{\circ} C$.

$(B)$ Final temperature of the system will be greater than $0^{\circ} C$.

$(C)$ 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:

A

$A$ and $D$ only

B

$B$ and $D$ only

C

$A$ and $E$ only

D

$A$ and $C$ only

(JEE MAIN-2023)

Solution

$\Delta Q =184 \times 10^3$

$m =0.600\,kg \text { at }-12^{\circ}\,C$

$S =222.3\,J / kg /{ }^{\circ}\,C$

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

$Q _1=0.600 \times 2222.3 \times 12=16000.56\,J$

Remaining heat $\Delta Q _1=184000-16000.56$

$=167999.44\,J$

For meeting at $0^{\circ}\,C$

$\Delta Q _2=0.600 \times 336000=201600\,J \text { needed }$

$\therefore 100 \%$ ice is not melted

Amount of ice melted

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

$\therefore$ mass of water $=0.4999\,kg$

Mass of ice $=0.1001$

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

Standard 11
Physics

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