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An ice cube of unknown mass at 0° C is added to 265 g of $H_{2} O$ at 25.00° C in a calorimeter. . If the final temperature of the resulting $H_{2} O$ is 21.70° C, what is the mass of the ice cube?
Properties of Water
$C_{p}$ $4.18 J \cdot g^{- 1} \cdot K^{- 1}$
$Δ H_{fusion}$ $333 J \cdot g^{- 1}$

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2.47 g

8.63 g

10.3 g

11.0 g

answer is B.

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Detailed Solution

$\begin{array}{l} m \times 333 + m \times 4.18 \times 21.7) \\ = 265 \times 4.18 \times (25 - 21.7) \\ \Rightarrow m = 8.63 g \end{array}$

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