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How much heat is required to convert 5.0 g of ice at -10.0° C to liquid water at 15.0° C? (Assume heat capacities are independent of temperature.)
Enthalpy of fusion $6.00 kJ \times {mol}^{- 1}$
Specific heat capacity of ice $37.8 J \times {mol}^{- 1} \times^{\circ} C^{- 1}$
Specific heat capacity of water $76.0 J \times {mol}^{- 1} \times^{\circ} C^{- 1}$

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$4.2 \times 10^{2} J$

$2.1 \times 10^{3} J$

$9.3 \times 10^{3} J$

$3.8 \times 10^{4} J$

answer is B.

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Detailed Solution

Required heat,

$\begin{array}{l} \frac{5}{18} \times 37.8 \times 10 + 6 \times 10^{3} \times \frac{5}{18} + 76 \times \frac{5}{18} \times 15 \\ = 2.1 \times 10^{3} J \end{array}$

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