Heat energy of 184,kJ is given to ice of mass | vedclass

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Question

(A) Given: Heat $\Delta Q = 184 	imes 10^3\,J$,mass $m = 0.6\,kg$,initial temperature $T_i = -12^{\circ}C$,specific heat $c_{ice} = 2222.3\,J\,kg^{-1

Vedclass · Accepted Answer

$A$ and $D$ only

Vedclass · Answer

(A) Given: Heat $\Delta Q = 184 	imes 10^3\,J$,mass $m = 0.6\,kg$,initial temperature $T_i = -12^{\circ}C$,specific heat $c_{ice} = 2222.3\,J\,kg^{-1\circ}C^{-1}$,latent heat $L = 336 	imes 10^3\,J\,kg^{-1}$.Step $1$: Heat required to raise the temperature of ice from $-12^{\circ}C$ to $0^{\circ}C$:$Q_1 = m \cdot c_{ice} \cdot \Delta T = 0.6 	imes 2222.3 	imes 12 = 16000.56\,J$.Step $2$: Remaining heat available for melting:$Q_{rem} = \Delta Q - Q_1 = 184000 - 16000.56 = 167999.44\,J$.Step $3$: Heat required to melt the entire mass of ice at $0^{\circ}C$:$Q_{melt} = m \cdot L = 0.6 	imes 336000 = 201600\,J$.Since $Q_{rem} Step $4$: Calculate the mass of ice melted $(m_w)$:$m_w = \frac{Q_{rem}}{L} = \frac{167999.44}{336000} \approx 0.5\,kg$.Step $5$: Calculate the remaining mass of ice $(m_i)$:$m_i = m - m_w = 0.6 - 0.5 = 0.1\,kg$.Step $6$: Ratio of ice to water:Ratio $= \frac{m_i}{m_w} = \frac{0.1}{0.5} = 1:5$. Thus,statement $(D)$ is correct.Therefore,$(A)$ and $(D)$ are correct.

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