When 0.15 ; kg of ice at 0^{circ} C is mixed | vedclass

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(D) Heat lost by water $= m_w s_w (	heta_i - 	heta_f) = (0.30 \; kg) (4186 \; J \; kg^{-1} \; K^{-1}) (50.0^{\circ} C - 6.7^{\circ} C) = 54376.14 \;

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$3.34 	imes 10^{5} \; J \; kg^{-1}$

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(D) Heat lost by water $= m_w s_w (	heta_i - 	heta_f) = (0.30 \; kg) (4186 \; J \; kg^{-1} \; K^{-1}) (50.0^{\circ} C - 6.7^{\circ} C) = 54376.14 \; J$.Heat gained by ice to melt $= m_i L_f = (0.15 \; kg) L_f$.Heat gained by melted ice to reach $6.7^{\circ} C = m_i s_w (	heta_f - 0^{\circ} C) = (0.15 \; kg) (4186 \; J \; kg^{-1} \; K^{-1}) (6.7^{\circ} C) = 4206.93 \; J$.By the principle of calorimetry,Heat lost = Heat gained:$54376.14 \; J = (0.15 \; kg) L_f + 4206.93 \; J$.$(0.15 \; kg) L_f = 54376.14 \; J - 4206.93 \; J = 50169.21 \; J$.$L_f = \frac{50169.21 \; J}{0.15 \; kg} = 3.3446 	imes 10^5 \; J \; kg^{-1} \approx 3.34 	imes 10^5 \; J \; kg^{-1}$.

When $0.15 \; kg$ of ice at $0^{\circ} C$ is mixed with $0.30 \; kg$ of water at $50^{\circ} C$ in a container,the resulting temperature is $6.7^{\circ} C$. Calculate the latent heat of fusion of ice. (Given: $s_{\text{water}} = 4186 \; J \; kg^{-1} \; K^{-1}$)

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