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Standard XII

Chemistry

Work Done in Isothermal Reversible Process

The molar hea...

Question

The molar heat capacity of water at constant pressure, $C_{p, m}$ is $75 J K^{- 1} m o l^{- 1}$ . when $1.0 k J$ of heat is supplied to $100 g$ of water which is free to expand, increase in the temperature of water is:

4.8 K

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6.6 K

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1.2 K

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2.4 K

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Solution

The correct option is D $2.4 K$
As, $C_{p, m} = \frac{q}{n (T_{2} - T_{1})} 75 = \frac{1000}{5.55 (Δ T)} (∵ n = \frac{100}{8} = 5.55) Δ T = \frac{1000}{5.55 \times 75} Δ T = 2.4 K$

Thus, the increase in temperature of water is $2.4 K$ .

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Work Done in Isothermal Reversible Process

Standard XII Chemistry

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The molar heat capacity of water at constant pressure, Cp,m is 75 JK−1mol−1. when 1.0 kJ of heat is supplied to 100 g of water which is free to expand, increase in the temperature of water is:

The correct option is D 2.4 K As, Cp,m=qn(T2−T1)75=10005.55(ΔT) (∵n=1008=5.55)ΔT=10005.55×75ΔT=2.4 K Thus, the increase in temperature of water is 2.4 K.

The molar heat capacity of water at constant pressure, $C_{p, m}$ is $75 J K^{- 1} m o l^{- 1}$ . when $1.0 k J$ of heat is supplied to $100 g$ of water which is free to expand, increase in the temperature of water is:

The correct option is D $2.4 K$
As, $C_{p, m} = \frac{q}{n (T_{2} - T_{1})} 75 = \frac{1000}{5.55 (Δ T)} (∵ n = \frac{100}{8} = 5.55) Δ T = \frac{1000}{5.55 \times 75} Δ T = 2.4 K$

Thus, the increase in temperature of water is $2.4 K$ .