When M_1 gram of ice at -10,^{circ}C (specifi | vedclass

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(A) According to the principle of calorimetry,$Heat\,lost = Heat\,gained$.Heat lost by $M_2$ grams of water cooling from $50\,^{\circ}C$ to $0\,^{\cir

Vedclass · Accepted Answer

$\frac{50M_2}{M_1} - 5$

Vedclass · Answer

(A) According to the principle of calorimetry,$Heat\,lost = Heat\,gained$.Heat lost by $M_2$ grams of water cooling from $50\,^{\circ}C$ to $0\,^{\circ}C$ is given by $Q_{lost} = M_2 	imes c_w 	imes \Delta T = M_2 	imes 1 	imes (50 - 0) = 50M_2$.Heat gained by $M_1$ grams of ice warming from $-10\,^{\circ}C$ to $0\,^{\circ}C$ and then melting at $0\,^{\circ}C$ is given by $Q_{gained} = M_1 	imes c_{ice} 	imes \Delta T + M_1 	imes L_f = M_1 	imes 0.5 	imes 10 + M_1 	imes L_f = 5M_1 + M_1 L_f$.Equating the two: $50M_2 = 5M_1 + M_1 L_f$.Rearranging for $L_f$: $M_1 L_f = 50M_2 - 5M_1$.Therefore,$L_f = \frac{50M_2}{M_1} - 5$.

When $M_1$ gram of ice at $-10\,^{\circ}C$ (specific heat $= 0.5\, cal\, g^{-1}\,^{\circ}C^{-1}$) is added to $M_2$ gram of water at $50\,^{\circ}C$,finally no ice is left and the water is at $0\,^{\circ}C$. The value of latent heat of ice,in $cal\, g^{-1}$ is

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