CBSE Class 11 Physics Chapter 12: Thermodynamics NCERT Solutions

The problem asks us to find the rate at which the fuel is consumed by a geyser, given the rate of water flow, the temperature change, and the heat of combustion of the fuel.

Given data:

• Rate of water flow = 3.0 litres per minute.
• Initial temperature of water, $T\_1\; =\; 27^\{\backslash circ\}\backslash text\{C\}$.
• Final temperature of water, $T\_2\; =\; 77^\{\backslash circ\}\backslash text\{C\}$.
• Heat of combustion of fuel = $4.0\; \backslash times\; 10^4\; \backslash text\{\; J/g\}$.
• Specific heat capacity of water, $c\; =\; 4.2\; \backslash text\{\; J\; g\}^\{-1\}\; \backslash ,^\{\backslash circ\}\backslash text\{C\}^\{-1\}$.

First, calculate the rise in temperature ($\backslash Delta\; T$):

$\backslash Delta\; T\; =\; T\_2\; -\; T\_1\; =\; 77^\{\backslash circ\}\backslash text\{C\}\; -\; 27^\{\backslash circ\}\backslash text\{C\}\; =\; 50^\{\backslash circ\}\backslash text\{C\}$

Next, determine the mass of water flowing per minute. Since the density of water is approximately 1 g/mL (or 1 kg/L), 3.0 litres of water is equal to 3000 grams.

Mass of flowing water per minute, $m\; =\; 3.0\; \backslash text\{\; L/min\}\; \backslash times\; 1000\; \backslash text\{\; g/L\}\; =\; 3000\; \backslash text\{\; g/min\}$

Now, calculate the total amount of heat required to raise the temperature of this mass of water using the formula $\backslash Delta\; Q\; =\; mc\; \backslash Delta\; T$:

$\backslash Delta\; Q\; =\; (3000\; \backslash text\{\; g\})\; \backslash times\; (4.2\; \backslash text\{\; J\; g\}^\{-1\}\; \backslash ,^\{\backslash circ\}\backslash text\{C\}^\{-1\})\; \backslash times\; (50^\{\backslash circ\}\backslash text\{C\})$

$\backslash Delta\; Q\; =\; 630000\; \backslash text\{\; J/min\}\; =\; 6.3\; \backslash times\; 10^5\; \backslash text\{\; J/min\}$

This is the rate at which heat energy must be supplied by the geyser.

The rate of consumption of fuel is found by dividing the total heat supplied per minute by the heat of combustion of the fuel:

Rate of fuel consumption = $\backslash frac\{\backslash text\{Total\; heat\; supplied\; per\; minute\}\}\{\backslash text\{Heat\; of\; combustion\}\}$

Rate of fuel consumption = $\backslash frac\{6.3\; \backslash times\; 10^5\; \backslash text\{\; J/min\}\}\{4.0\; \backslash times\; 10^4\; \backslash text\{\; J/g\}\}$

Rate of fuel consumption = $15.75\; \backslash text\{\; g/min\}$

Therefore, the rate of consumption of the fuel is 15.75 grams per minute.

This question requires us to calculate the heat energy needed to increase the temperature of a specific amount of nitrogen gas at constant pressure.

Given data:

• Mass of nitrogen, $m\; =\; 2.0\; \backslash times\; 10^\{-2\}\; \backslash text\{\; kg\}\; =\; 20\; \backslash text\{\; g\}$.
• Rise in temperature, $\backslash Delta\; T\; =\; 45^\{\backslash circ\}\backslash text\{C\}$. Since the change in temperature is the same in Celsius and Kelvin, $\backslash Delta\; T\; =\; 45\; \backslash text\{\; K\}$.
• Molecular mass of $N\_2$, $M\; =\; 28\; \backslash text\{\; g/mol\}$.
• Universal gas constant, $R\; =\; 8.3\; \backslash text\{\; J\; mol\}^\{-1\}\; \backslash text\{\; K\}^\{-1\}$.

First, calculate the number of moles ($n$) of nitrogen:

$n\; =\; \backslash frac\{\backslash text\{mass\}\}\{\backslash text\{Molecular\; mass\}\}\; =\; \backslash frac\{m\}\{M\}$

$n\; =\; \backslash frac\{20\; \backslash text\{\; g\}\}\{28\; \backslash text\{\; g/mol\}\}\; \backslash approx\; 0.714\; \backslash text\{\; mol\}$

Nitrogen ($N\_2$) is a diatomic gas. For a diatomic gas, the molar specific heat at constant pressure ($C\_P$) is given by:

$C\_P\; =\; \backslash frac\{7\}\{2\}R$

Substitute the value of R:

$C\_P\; =\; \backslash frac\{7\}\{2\}\; \backslash times\; 8.3\; \backslash text\{\; J\; mol\}^\{-1\}\; \backslash text\{\; K\}^\{-1\}$

$C\_P\; \backslash approx\; 29.05\; \backslash text\{\; J\; mol\}^\{-1\}\; \backslash text\{\; K\}^\{-1\}$

The amount of heat ($\backslash Delta\; Q$) to be supplied at constant pressure is given by the formula:

$$\backslash Delta\; Q\; =\; nC\_P\; \backslash Delta\; T$$

Now, substitute the values:

$\backslash Delta\; Q\; =\; (0.714\; \backslash text\{\; mol\})\; \backslash times\; (29.05\; \backslash text\{\; J\; mol\}^\{-1\}\; \backslash text\{\; K\}^\{-1\})\; \backslash times\; (45\; \backslash text\{\; K\})$

$\backslash Delta\; Q\; \backslash approx\; 933.38\; \backslash text\{\; J\}$

Therefore, the amount of heat that must be supplied to 2.0 x 10^-2 kg of nitrogen to raise its temperature by 45 °C at constant pressure is approximately 933.38 Joules.