How much energy must be supplied to change 36 g of ice at 0 ° C to water at room temperature 25 ° C ? Data for water Δ H f u s i o n o 6 .01 kJ/mol C p (liquid) 4 .18 JK − 1 g − 1

Δ H = ( Δ H f u s i o n ) of two moles + m s Δ t = 6.01 × 2 + 36 × 4.18 × 25 × 1 0 - 3 = 16 kJ.

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ChemistryThermochemistryEnthalpy of ReactionEasy2 minPYQ_2020

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How much energy must be supplied to change 36 g of ice at $0 ° C$ to water at room temperature $25 ° C$ ?

Data for water
$Δ H_{f u s i o n}^{o}$	$6 .01 kJ/mol$
$C_{p (liquid)}$	$4 .18 {JK}^{- 1} g^{- 1}$

Answer:

Solution:

$Δ H = (Δ H_{f u s i o n})$ of two moles $+ m s Δ t$
$= 6.01 \times 2 + 36 \times 4.18 \times 25 \times 1 0^{- 3}$
= 16 kJ.

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:ThermochemistrySubtopic:Enthalpy of Reaction

⏱ 2mℹ️ Source: PYQ_2020

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