Question 270: Thermodynamics & KTG

Question

$$1 \mathrm{~kg}$$ of water at $$100^{\circ} \mathrm{C}$$ is converted into steam at $$100^{\circ} \mathrm{C}$$ by boiling at atmospheric pressure. The volume of water changes from $$1.00 	imes 10^{-3} \mathrm{~m}^{3}$$ as a liquid to $$1.671 \mathrm{~m}^{3}$$ as steam. The change in internal energy of the system during the process will be
(Given latent heat of vaporisaiton $$=2257 \mathrm{~kJ} / \mathrm{kg}$$, Atmospheric pressure = $$\left.1 	imes 10^{5} \mathrm{~Pa}ight)$$

Accepted Answer

To find the change in internal energy, we need to consider both the heat added during the process and the work done during the process.
First, let's calculate the heat added ($Q$) to convert 1 kg of water at 100ÃÂ°C into steam at 100ÃÂ°C using the latent heat of vaporization:
$$Q = m 	imes L$$
where $$m = 1 \mathrm{~kg}$$ and $$L = 2257 \mathrm{~kJ/kg}$$:
$$Q = 1 \mathrm{~kg} 	imes 2257 \mathrm{~kJ/kg} = 2257 \mathrm{~kJ}$$
Next, let's calculate the work done ($W$) on the system during the process. The work done is given by:
$$W = -P \Delta V$$
where $$P$$ is the atmospheric pressure and $$\Delta V$$ is the change in volume. We are given that the atmospheric pressure is $$P = 1 	imes 10^5 \mathrm{~Pa}$$, and the change in volume is
$$\Delta V = 1.671 \mathrm{~m}^3 - 1.00 	imes 10^{-3} \mathrm{~m}^3 = 1.670 \mathrm{~m}^3$$.
Now, we can calculate the work done:
$$W = -(1 	imes 10^5 \mathrm{~Pa})(1.670 \mathrm{~m}^3) = -167000 \mathrm{~J} = -167 \mathrm{~kJ}$$
Finally, we can find the change in internal energy ($\Delta U$) using the first law of thermodynamics:
$$\Delta U = Q + W$$
Substitute the values of $$Q$$ and $$W$$:
$$\Delta U = 2257 \mathrm{~kJ} - 167 \mathrm{~kJ} = 2090 \mathrm{~kJ}$$
The change in internal energy of the system during the process is +2090 kJ.

Answer

+ 2090 kJ

Answer

$$-$$ 2426 kJ

Answer

+ 2476 kJ

Answer

$$-$$ 2090 kJ

Thermodynamics & KTG

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