An ice cube at an unknown temperature is added to 25.0 g of liquid $H_{2} O$ at 40.0° C. The final temperature of the 29 .3 g equilibrated mixture is 21.5° C. What was the original temperature of the ice cube? $[C_{p} (J / g \times^{\circ} C)$ water= 4.184, ice= 2.06, $Δ H_{fusion}^{\circ} = 333 J / g]$ .

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-6.5° C

-13.1° C

-35.3° C

-56.8° C

answer is B.

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Detailed Solution

Original temperature of the ice cube is given as,

$\begin{array}{l} (4.3 \times 2.06 \times T + 333 \times 4.3 + 4.3 \times 4.184 \times 21.5) \\ = (25 \times 4.184 \times (40 - 21.5) \\ \Rightarrow T = - {13.1}^{\circ} C \end{array}$

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