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Thermodynamics

Section: Chemistry

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Question : 2 of 34

Marks: +1, -0

The

C_{p}

\mathrm{H}_2\mathrm{O}(l)

75.3\ \mathrm{J\,mol^{-1}\,K^{-1}}

. What is the energy (in J ) required to raise 180 g of liquid water from

10^{\circ}\mathrm{C}

15^{\circ}\mathrm{C} ? (\mathrm{H}_2\mathrm{O}=18\ \mathrm{u})

[AP EAPCET 24 May 2025 S1]

$3.765$
$3765$
$753$
$376.5$

Solution:

First, we need to find out how many moles of water we have.

n\;=\;\frac{m}{M}\;=\;\frac{180}{18}

\;=10\ \mathrm{mol}

\Delta T\;=T_2-T_1=15-10=5\ \mathrm{K}

The energy needed to heat the water is given by the formula:

Q=n C_p \Delta T

Here,

n

is number of moles (10),

C_p

is the specific heat

(75.3\ \mathrm{J\,mol^{-1}\,K^{-1}})

, and

\Delta T

is the temperature change (5 K).

So,

Q=10 \times 75.3 \times 5=3765\ \mathrm{J}

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