The molar heat capacity of water at constant, pressure, C , is 75 ~J ~K^ -1 ~mol^ -1 . When 1.0 kJ of heat is supplied to 100 ~g of water which is free to expand, the increase in temperature of water is:

Correct answer: B. 2.4 ~K Molar heat capacity =75 J K^ -1 mol^ -1 aligned & 18 g of water =1 mole =75 J K^ -1 & mol^ -1 & 1 g of water =(75)/(18) JK^ -1 & Q= m.c. Δ t aligned aligned 1000 & =100 × (75)/(18) × Δ t Δ t & =(10 × 18)/(75)=2.4 K aligned

NEET 2003

Exam: NEET
Year: 2003
Subject: Chemistry
Chapter: Thermodynamics (C)

NEET 2003 · Chemistry · Thermodynamics (C)

The molar heat capacity of water at constant, pressure, C , is 75...

The molar heat capacity of water at constant, pressure,

\mathrm{C}

, is

75 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

. When 1.0

\mathrm{kJ}

of heat is supplied to

100 \mathrm{~g}

of water which is free to expand, the increase in temperature of water is:

Options

A $1.2 \mathrm{~K}$
B $2.4 \mathrm{~K}$
C $4.8 \mathrm{~K}$
D $6.6 \mathrm{~K}$

Correct answer

B. $2.4 \mathrm{~K}$

Step-by-step solution

Molar heat capacity

=75 \mathrm{ J} \mathrm{ K}^{-1} \mathrm{ mol}^{-1}

\begin{aligned} & 18 \mathrm{ g} \text{ of water }=1 \text{ mole }=75 \mathrm{ J} \mathrm{ K}^{-1} \\ & \mathrm{ mol}^{-1} \\ & 1 \mathrm{ g} \text{ of water }=\frac{75}{18} \mathrm{JK}^{-1} \\ & \mathrm{Q}=\text{ m.c. } \Delta t \end{aligned}

\begin{aligned} 1000 & =100 \times \frac{75}{18} \times \Delta t \\ \Rightarrow \Delta t & =\frac{10 \times 18}{75}=2.4 \mathrm{ K} \end{aligned}

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