Question 176: Thermodynamics & KTG

Question

A gas based geyser heats water flowing at the rate of 5.0 litres per minute from $27^{\circ} \mathrm{C}$ to $87^{\circ} \mathrm{C}$. The rate of consumption of the gas is $\_\_\_\_$ $\mathrm{g} / \mathrm{s}$.
(Take heat of combustion of gas $=5.0 	imes 10^4 \mathrm{~J} / \mathrm{g}$ ) specific heat capacity of water $=4200 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}$

Accepted Answer

The heat energy ( Q ) required to raise the temperature of a mass ( m ) is given by $\mathrm{Q}=\mathrm{mc} \Delta \mathrm{T}$
Where c is the specific heat capacity and $\Delta \mathrm{T}$ is the change in temperature.
Temperature change is $\Delta \mathrm{T}=87^{\circ} \mathrm{C}-27^{\circ} \mathrm{C}=60^{\circ} \mathrm{C}$
Specific heat is $\mathrm{c}=4200 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}$.
The heat required per unit time is $\mathrm{P}_{\mathrm{req}}=\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{dm}}{\mathrm{dt}} \cdot \mathrm{c} \cdot \Delta \mathrm{T}$
Water flows at 5.0 litres $/ \min$.
Since the density of water is $1 \mathrm{~kg} / \mathrm{L}$, the mass flow rate is $5.0=\frac{5.0}{60} \mathrm{~kg} / \mathrm{sec}$
$$ P_{\mathrm{req}}=\left(\frac{5.0}{60}ight) 	imes 4200 	imes 60 \frac{\mathrm{~J}}{\mathrm{~s}}=21000 \mathrm{~J} / \mathrm{s} $$
Let the rate of consumption of gas be $\mathrm{R} \mathrm{g} / \mathrm{s}$.
The heat produced per second is,
$$ \mathrm{P}_{	ext {prod }}=\mathrm{R} 	imes \mathrm{H} $$
Given heat of combustion of gas $\mathrm{H}=5.0 	imes 10^4 \mathrm{~J} / \mathrm{g}$
$$ \mathrm{P}_{	ext {prod }}=\mathrm{R} 	imes 50000 \mathrm{~J} / \mathrm{s} $$
Assuming 100% efficiency (because no heat loss is given), $\mathrm{P}_{	ext {prod }}=\mathrm{P}_{	ext {req }}$
$$ R 	imes 50000=21000 $$
$\Rightarrow \mathrm{R}=\frac{21000}{50000}=0.42 \mathrm{~g} / \mathrm{s}$
Therefore, the gas must be consumed at a rate of $0.42 \mathrm{~g} / \mathrm{s}$ to maintain the temperature rise.
Hence, the correct option is (D).

Answer

4.2

Answer

2.1

Answer

0.21

Answer

0.42

Thermodynamics & KTG

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