Question

Assume that for a domestic hot water supply 150Kg of water per day must be heated from ${10^o}C$ to ${65^o}C$ and gaseous fuel propane ${C_3}{H_8}$ is used for this purpose. The volume of propane (in $litre$ at $STP$) would have to be used for heating domestic water is:
Given : $\Delta H$ for combustion of propane is $ - 2050kJmo{l^{ - 1}}\;$and specific heat of water is $4.184 \times {10^{ - 3}}kJ/g$
A. $1.22 \times {10^2}litre$
B. $7.55 \times {10^2}litre$
C. $2.11 \times {10^2}litre$
D. $3.77 \times {10^2}litre$

Answer 1

Hint: The specific heat mostly varies with temperature, and with each state of matter, it is different. Water has a high specific heat potential meaning it requires more energy to increase the temperature of water relative to other liquids. Specific heat of gases have differences to a higher extent when compared to other states of matter.

Formula used:
$Q = mC\Delta t$
$Q{\text{ }} = $quantity of heat absorbed
$m{\text{ }} = $mass of the body
$\Delta t{\text{ }} = $Increase in temperature
$C{\text{ }} = $Specific heat capacity of a substa

Complete step by step answer:
Let us first understand what is specific heat;
A substance 's heat capacity is the amount of heat needed to increase the entire substance's temperature by one degree. If the mass of the material is unity then it is called specific heat capacity or specific heat.
Now, let us move into the calculations;
Let us analyse the given data;
$m = 150Kg = 15000g$ which is the mass here
$\Delta t = 65 - 10 = {55^o}C$ which is the change in temperature
$C = 4.184 \times {10^{ - 3}}$ which is the specific heat capacity of water.
Therefore by substituting the above values we get;
$Q = mC\Delta t = 15000 \times 55 \times 4.184 \times {10^{ - 3}} = 34518KJ$
It is given that $2050KJ$ heat is provided by one mole of propane ${C_3}{H_8}$
So, we have to find out how many moles of propane produced $34518KJ$ heat, for this we just have to divide the two quantities
Number of moles $ = \dfrac{{34518}}{{2050}} = 16.838mol$
Then to find out the volume we can use the formula of number of moles;
$n = \dfrac{\text{given volume}}{{22.4l}}$ where, $n$ is the number of moles$ = 16.838mol$ and $22.4l = $ volume at STP.
Therefore Volume $ = 22.4 \times 16.838 = 3.77 \times {10^2}litres$

So, Option D is correct.

Note: The water temperature, as in most liquids, increases as it absorbs heat and decreases as it loses heat. The temperature of water, however, decreases and grows more slowly than most other liquids. Without an immediate temperature increase, we may assume that water absorbs heat. Its temperature is therefore maintained for longer than other compounds.