Total enthalpy change for freezing of 1 mol of water at $1 0^{\circ} C$ to ice at $- 1 0^{\circ} C$ is ____
(Given : $Δ_{fus} H = x kJ / mol$ )
$C_{p} [H_{2} O (l)] = yJ mol^{- 1} K^{- 1} C_{p} [H_{2} O (s)] = zJ mol^{- 1} K^{- 1}$

Question

Question diagram: Total enthalpy change for freezing of 1 mol of water at 1 0

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Answer:

B

Solution:

$Δ H = 1 \times y (0 - 10) - x \times 1000 + 1 \times z (- 1 0^{\circ} - 0^{\circ}) Δ H = - 10 (100 x + y + z) Joule.$

TestHub · Accepted Answer

​ Δ H = 1 × y ( 0 − 10 ) − x × 1000 + 1 × z ( − 1 0 ∘ − 0 ∘ ) Δ H = − 10 ( 100 x + y + z ) Joule. ​

Chemistry - Thermochemistry Question with Solution | TestHub

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