Question

The molar heat capacity of water at constant pressure ${C_p}$ is $75J{K^{ - 1}}mo{l^{ - 1}}$ . When $10kJ$ of heat is supplied to $1kg{H_2}O$ which is free to expand, the increase in temperature of ${H_2}O$ is:
A. $2.4K$
B. $4.8K$
C. $3.2K$
D. $10K$

Answer 1

The amount of heat added so as to increase the temperature by one unit is called molar heat capacity. There is a relationship between the heat supplied, molar heat capacity at constant pressure and temperature change. So, if we apply heat to any system there will be an increment in the temperature of the system.

Formula used:
$q = n \times {C_p} \times \Delta T$
Where, $q$ = heat
${C_p}$ = heat capacity at constant pressure
$n$ = number of moles
$\Delta T$ = Change in Temperature

Complete step by step answer:
Heat capacity is the heat needed to raise the temperature of a system by ${1^0}C$ under a given process.
Mathematically it is represented as,
$C = \dfrac{{dq}}{{dT}}J{C^{ - 1}}$
Molar heat capacity is the heat required to raise the temperature of 1 mole of the substance by unit degree.
Mathematically it is represented as,
$C = \dfrac{{dq}}{{ndT}}J{K^{ - 1}}mo{l^{ - 1}}$
Specific heat capacity is the heat required to raise the temperature of unit mass of a substance by unit degree temperature.
Mathematically it is represented as,
$s = \dfrac{{dq}}{{mdT}}J{g^{ - 1}}{K^{ - 1}}$
So, in this question we need molar heat capacity,
Therefore,
$C = \dfrac{{dq}}{{ndT}}$
Now,
$dq = nCdT$
Let’s, integrate the above equation
$q = \smallint nCdT$
Now we get,
$q = n \times {C_p} \times \Delta T$
The relationship between the heat supplied, molar heat capacity at constant pressure and temperature change will be given as follows,
$q = n \times {C_p} \times \Delta T$
The given values are,
${C_p} = 75J{K^{ - 1}}mo{l^{ - 1}}$
$q = 10kJ$
So, the heat,
$q = 10kJ = 10000J$
Weight of the water,
$Mass = 1kg = 1000g$
Change in the temperature we have to find,
i.e., $\Delta T = ?$
Number of moles,
$n = \dfrac{{{\text{mass}}}}{{{\text{molecular mass}}}}$
The molecular weight of water = $18gmo{l^{ - 1}}$
So,
$n = \dfrac{{1000}}{{18}}$
Let’s substitute the values in the above expression,
$10000 = \dfrac{{1000}}{{18}} \times 75 \times \Delta T$
Then we get,
$\Delta T = \dfrac{{10000 \times 18}}{{75 \times 1000}}$
Now we obtain the change in temperature,
$\Delta T = 2.4K$
Hence, the change in temperature will be $2.4K$ .

So, the correct answer is Option A.

Note: Do make sure that the given values are converted in the same unit. In simpler words, all quantities should be converted in the same unit while solving the answer. Since, the molar heat capacity is given in the terms of $J{K^{ - 1}}mo{l^{ - 1}}$ therefore, the heat is given in $kJ$ so make sure that it is converted into the $J$ .