Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The temperature of 2 kg of neon is increased from 20 to \(180^{\circ} \mathrm{C}\). Calculate the change in the total internal energy of the neon, in \(\mathrm{kJ}\). Would the internal energy change be any different if the neon were replaced with argon?

Short Answer

Expert verified
Would the internal energy change be different if neon were replaced with argon?

Step by step solution

01

Convert temperature to Kelvin.

First, we need to convert the given temperatures from Celsius to Kelvin, as the calculations should be performed in Kelvin. K = °C + 273.15 Initial temperature (T1): 20°C = 293.15 K Final temperature (T2): 180°C = 453.15 K
02

Calculate the change in temperature (ΔT)

Now, we will calculate the change in temperature. ΔT = T2 - T1 = 453.15 K - 293.15 K = 160 K
03

Find the specific heat capacity and mass

We are given that the mass (m) of the neon is 2 kg. The specific heat capacity (c) of neon is 1.03 J/g*K (joules per gram per Kelvin). Since we have the mass in kilograms, we will convert the specific heat capacity to J/kg*K (joules per kilogram per Kelvin). 1 kg = 1000 g c = 1.03 J/g*K × 1000 g/kg = 1030 J/kg*K
04

Calculate the change in internal energy (ΔU) for neon

Now, we will use the formula: ΔU = mcΔT ΔU = (2 kg) × (1030 J/kg*K) × (160 K) ΔU = 328960 J Since we need the answer in kJ, we will convert from J to kJ. 1 kJ = 1000 J ΔU = 328960 J ÷ 1000 = 328.96 kJ The change in total internal energy of the neon is 328.96 kJ.
05

Evaluate the internal energy change for argon

To find out if the internal energy change would be different for argon, we need to know the specific heat capacity of argon. The specific heat capacity of argon is 0.520 J/g*K, which is different from neon's specific heat capacity. This means that the internal energy change for argon would also be different. In conclusion, the change in the total internal energy of 2 kg of neon when its temperature is increased from 293.15 K to 453.15 K is 328.96 kJ. If neon were replaced with argon, the internal energy change would be different.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A room contains 75 kg of air at 100 kPa and \(15^{\circ} \mathrm{C}\) The room has a 250 -W refrigerator (the refrigerator consumes \(250 \mathrm{W} \text { of electricity when running }),\) a \(120-\mathrm{W} \mathrm{TV},\) a 1.8-kW electric resistance heater, and a 50-W fan. During a cold winter day, it is observed that the refrigerator, the TV, the fan, and the electric resistance heater are running continuously but the air temperature in the room remains constant. The rate of heat loss from the room that day is \((a) 5832 \mathrm{kJ} / \mathrm{h}\) (b) \(6192 \mathrm{kJ} / \mathrm{h}\) \((c) 7560 \mathrm{kJ} / \mathrm{h}\) \((d) 7632 \mathrm{kJ} / \mathrm{h}\) \((e) 7992 \mathrm{kJ} / \mathrm{h}\)

Stainless steel ball bearings \(\left(\rho=8085 \mathrm{kg} / \mathrm{m}^{3}\right.\) and \(\left.c_{p}=0.480 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) having a diameter of \(1.2 \mathrm{cm}\) are to be quenched in water at a rate of 800 per minute. The balls leave the oven at a uniform temperature of \(900^{\circ} \mathrm{C}\) and are exposed to air at \(25^{\circ} \mathrm{C}\) for a while before they are dropped into the water. If the temperature of the balls drops to \(850^{\circ} \mathrm{C}\) prior to quenching, determine the rate of heat transfer from the balls to the air.

A piston-cylinder device initially contains \(0.4 \mathrm{kg}\) of nitrogen gas at \(160 \mathrm{kPa}\) and \(140^{\circ} \mathrm{C}\). The nitrogen is now expanded isothermally to a pressure of 100 kPa. Determine the boundary work done during this process.

A \(0.3-\mathrm{L}\) glass of water at \(20^{\circ} \mathrm{C}\) is to be cooled with ice to \(5^{\circ} \mathrm{C}\). Determine how much ice needs to be added to the water, in grams, if the ice is at \((a) 0^{\circ} \mathrm{C}\) and \((b)-20^{\circ} \mathrm{C}\) Also determine how much water would be needed if the cooling is to be done with cold water at \(0^{\circ} \mathrm{C}\). The melting temperature and the heat of fusion of ice at atmospheric pressure are \(0^{\circ} \mathrm{C}\) and \(333.7 \mathrm{kJ} / \mathrm{kg}\), respectively, and the density of water is \(1 \mathrm{kg} / \mathrm{L}\).

A 2 -kW electric resistance heater submerged in \(5-\mathrm{kg}\) water is turned on and kept on for 10 min. During the process, \(300 \mathrm{kJ}\) of heat is lost from the water. The temperature rise of water is \((a) 0.4^{\circ} \mathrm{C}\) (b) \(43.1^{\circ} \mathrm{C}\) \((c) 57.4^{\circ} \mathrm{C}\) \((d) 71.8^{\circ} \mathrm{C}\) \((e) 180^{\circ} \mathrm{C}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free