300 , g of water at 25^{circ} C is added to 1 | vedclass

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(D) Heat lost by water to cool down to $0^{\circ} C$ is given by $Q_{lost} = m_w c_w \Delta T = 300 	imes 1 	imes (25 - 0) = 7500 \, cal$.Heat requi

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(D) Heat lost by water to cool down to $0^{\circ} C$ is given by $Q_{lost} = m_w c_w \Delta T = 300 	imes 1 	imes (25 - 0) = 7500 \, cal$.Heat required to melt $100 \, g$ of ice at $0^{\circ} C$ is $Q_{gain} = m_i L_f = 100 	imes 80 = 8000 \, cal$ (taking $L_f \approx 80 \, cal/g$).Since $Q_{lost} Specifically,the amount of ice that melts is $m_{melted} = \frac{Q_{lost}}{L_f} = \frac{7500}{80} = 93.75 \, g$.Since there is still some ice remaining $(100 - 93.75 = 6.25 \, g)$ in the mixture,the system remains at the phase equilibrium temperature of $0^{\circ} C$.Therefore,the final temperature of the mixture is $0^{\circ} C$.

$300 \, g$ of water at $25^{\circ} C$ is added to $100 \, g$ of ice at $0^{\circ} C$. The final temperature of the mixture is ........... $^{\circ} C$.

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