X , g of ice at 0^{circ} C is added to 340 , | vedclass

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(D) The heat gained by the ice equals the heat lost by the water.Heat gained by $X \, g$ of ice at $0^{\circ} C$ to become water at $5^{\circ} C$ is g

Vedclass · Accepted Answer

$60.3$

Vedclass · Answer

(D) The heat gained by the ice equals the heat lost by the water.Heat gained by $X \, g$ of ice at $0^{\circ} C$ to become water at $5^{\circ} C$ is given by: $Q_{gain} = X 	imes L_f + X 	imes c 	imes \Delta T_{ice}$.$Q_{gain} = X(333) + X(4.184)(5 - 0) = 333X + 20.92X = 353.92X$.Heat lost by $340 \, g$ of water cooling from $20^{\circ} C$ to $5^{\circ} C$ is given by: $Q_{lost} = m 	imes c 	imes \Delta T_{water}$.$Q_{lost} = 340 	imes 4.184 	imes (20 - 5) = 340 	imes 4.184 	imes 15 = 21338.4 \, J$.Equating $Q_{gain} = Q_{lost}$:$353.92X = 21338.4$.$X = \frac{21338.4}{353.92} \approx 60.29 \, g$.Thus,the value of $X$ is approximately $60.3 \, g$.

$X \, g$ of ice at $0^{\circ} C$ is added to $340 \, g$ of water at $20^{\circ} C$. The final temperature of the resultant mixture is $5^{\circ} C$. The value of $X$ (in $g$) is closest to:
[Heat of fusion of ice $= 333 \, J / g$; specific heat of water $= 4.184 \, J / g \cdot K$]

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$X \, g$ of ice at $0^{\circ} C$ is added to $340 \, g$ of water at $20^{\circ} C$. The final temperature of the resultant mixture is $5^{\circ} C$. The value of $X$ (in $g$) is closest to:[Heat of fusion of ice $= 333 \, J / g$; specific heat of water $= 4.184 \, J / g \cdot K$]