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Chapter 8: Problem 1118

If heat given to a system is \(6 \mathrm{k}\) cal and work done is $6 \mathrm{kj}$. The change in internal energy is ......... KJ. (A) \(12.4\) (B) 25 (C) \(19.1\) (D) 0

Short Answer

Expert verified
The change in internal energy is 19.1 KJ.

Step by step solution

01

Understand the problem and identify what is given

In this system, heat and work are being added to the system. The heat added to the system is 6 kcal and the work done on the system is 6 kJ. You need to find the change in internal energy, which is the sum of the heat added and the work done.
02

Convert the units of heat or work to be the same

For consistency and to be able to add the two forms of energy together, they must be in the same units. We are asked for the answer in KJ, so convert calories to Joules. There are 4.184 Joules in a calorie, and because the heat is given in kcals, multiply by 1000 to convert to kJ. Thus, 6kcal equates to 6*4.184 = 25.104 kJ.
03

Calculate the change in internal energy

To calculate the change in internal energy, simply add the heat added (which was converted to kJ in step 2) to the work done because the First Law of Thermodynamics states that the internal energy change of a system is equal to the amount of heat added to the system, plus the work done on it. So, 25.104 kJ (from heat) + 6 kJ (from work) equals 31.104 kJ. However, the answer options do not contain 31.104 KJ, which may suggest that the exercise may consider that the work done to the system is actually done by the system, therefore work done should be subtracted from the heat absorbed.
04

Recalculate the change in internal energy by subtracting the work.

If work is done by the system it means that energy was used by the system. Consequently, to calculate the accurate change in internal energy subtract the work from the heat: 25.104 kJ (from heat) - 6 kJ (from work) equals 19.104 kJ. The answer closest to this result given in the options is (C) 19.1 kJ. Therefore, this would be the correct answer in this case.

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Most popular questions from this chapter

In Column I different Process is given match corresponding option of column \(\mathrm{I}_{1}\) Column - I Columm - II (A) adiabatic process (p) \(\Delta \mathrm{p}=0\) (B) Isobaric process (q) \(\Delta \mathrm{u}=0\) (C) Isochroic process (r) \(\Delta Q=0\) (D) Isothermal process (s) \(\Delta \mathrm{W}=0\) (A) A-p, B-s, C-r, D-q (B) (B) A-s, B-q, C-p, D-r (C) $\mathrm{A}-\mathrm{r}, \mathrm{B}-\mathrm{p}, \mathrm{C}-\mathrm{s}, \mathrm{D}-\mathrm{q}$ (D) $\mathrm{A}-\mathrm{q}, \mathrm{B}-\mathrm{r}, \mathrm{C}-\mathrm{q}, \mathrm{D}-\mathrm{p}$

A System under goes a Cyclic Process in which it absorbs \(\mathrm{Q}_{1}\) heat and gives out \(\mathrm{Q}_{2}\) heat. The efficiency of the Process \(\eta\) and the work done is \(\mathrm{W}\). Which formula is wrong ? is (A) \(\mathrm{W}=\mathrm{Q}_{1}-\mathrm{Q}_{2}\) (B) \(\eta=\left(\mathrm{Q}_{2} / \mathrm{Q}_{1}\right)\) (C) \(\eta=\left(\mathrm{W} / \mathrm{Q}_{1}\right)\) (D) \(\eta=1-\left(\mathrm{Q}_{2} / \mathrm{Q}_{1}\right)\)

If a quantity of heat \(1163.4 \mathrm{~J}\) is supplied to one mole of nitrogen gas, at room temperature at constant pressure, then the rise in temperature is \(\mathrm{R}=8.31\\{\mathrm{~J} /(\mathrm{m} \cdot 1 . \mathrm{k})\\}\) (A) \(28 \mathrm{~K}\) (B) \(65 \mathrm{~K}\) (C) \(54 \mathrm{~K}\) (D) \(40 \mathrm{~K}\)

For an isothermal expansion of a Perfect gas, the value of $(\Delta \mathrm{P} / \mathrm{P})$ is equal to (A) \(-\gamma^{(1 / 2)}\\{(\Delta \mathrm{V}) / \mathrm{V}\\}\) (B) \(-\gamma\\{(\Delta \mathrm{V}) / \mathrm{V}\\}\) (C) \(-\gamma^{2}\\{(\Delta \mathrm{V}) / \mathrm{V}\\}\) \((\mathrm{D})-(\Delta \mathrm{V} / \mathrm{V})\)

One mole of an ideal gas $\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)=\gamma$ at absolute temperature \(\mathrm{T}_{1}\) is adiabatically compressed from an initial pressure \(\mathrm{P}_{1}\) to a final pressure \(\mathrm{P}_{2}\) The resulting temperature \(\mathrm{T}_{2}\) of the gas is given by. (A) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\\{\gamma /(\gamma-1)\\}}$ (B) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\\{(\gamma-1) / \gamma\\}}$ (C) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\gamma}$ (D) $\mathrm{T}_{2}=\mathrm{T}_{1}\left(\mathrm{p}_{2} / \mathrm{p}_{1}\right)^{\gamma-1}$
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