10; g of ice cubes at 0;^{circ} C are release | vedclass

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(A) Let the final temperature be $T$.Heat gained by ice to melt and reach temperature $T$ is given by $Q_{gain} = m_i L_f + m_i c_w (T - 0)$.Substitut

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$22$

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(A) Let the final temperature be $T$.Heat gained by ice to melt and reach temperature $T$ is given by $Q_{gain} = m_i L_f + m_i c_w (T - 0)$.Substituting the values: $Q_{gain} = 10 	imes 80 + 10 	imes 1 	imes T = 800 + 10T$.Heat lost by the water and the tumbler (water equivalent $55\; g$) is given by $Q_{lost} = (m_w + W) c_w (40 - T)$.Here,the water equivalent $W = 55\; g$ acts as additional mass of water. Assuming the initial mass of water in the tumbler is $55\; g$ (as implied by the water equivalent context),$Q_{lost} = 55 	imes 1 	imes (40 - T)$.By the law of calorimetry,$Q_{gain} = Q_{lost}$.$800 + 10T = 55(40 - T)$.$800 + 10T = 2200 - 55T$.$65T = 1400$.$T = 1400 / 65 \approx 21.54^{\circ} C$.Rounding to the nearest integer,$T \approx 22^{\circ} C$.

$10\; g$ of ice cubes at $0\;^{\circ} C$ are released in a tumbler (water equivalent $55\; g$) at $40\;^{\circ} C$. Assuming that negligible heat is taken from the surroundings,the temperature (in $^{\circ} C$) of water in the tumbler becomes nearly $(L_f = 80\; cal/g)$.

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