How much energy in kJ is required to convert | vedclass

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(A) Step $1$: Calculate the moles of water: $n = \frac{54 \ g}{18 \ g \ mol^{-1}} = 3 \ mol$. Step $2$: Calculate the heat required for melting ice at

Vedclass · Accepted Answer

$24.12$

Vedclass · Answer

(A) Step $1$: Calculate the moles of water: $n = \frac{54 \ g}{18 \ g \ mol^{-1}} = 3 \ mol$. Step $2$: Calculate the heat required for melting ice at $0 \ ^\circ C$ $(q_1)$: $q_1 = n 	imes \Delta H_{fusion} = 3 \ mol 	imes 6.01 \ kJ \ mol^{-1} = 18.03 \ kJ$. Step $3$: Calculate the heat required to raise the temperature of water from $0 \ ^\circ C$ to $27 \ ^\circ C$ $(q_2)$: $q_2 = m 	imes C_p 	imes \Delta T = 54 \ g 	imes 4.18 \ J \ K^{-1} \ g^{-1} 	imes 27 \ K = 6094.44 \ J = 6.09 \ kJ$. Step $4$: Total energy required = $q_1 + q_2 = 18.03 \ kJ + 6.09 \ kJ = 24.12 \ kJ$.

How much energy in $kJ$ is required to convert $54 \ g$ of ice at $0 \ ^\circ C$ to water at $27 \ ^\circ C$? $\left( \Delta H_{fusion} = 6.01 \ kJ \ mol^{-1}, C_{p(liquid)} = 4.18 \ J \ K^{-1} \ g^{-1} \right)$

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