GO An insulated Thermos contains 130 g of water at 80.0°C. You put in a 12.0 g ice cube at 0°C to form a system of ice + original water. (a) What is the equilibrium temperature of the system? What are the entropy changes of the water that was originally the ice cube (b) as it melts and (c) as it warms to the equilibrium temperature? (d) What is the entropy change of the original water as it cools to the equilibrium temperature? (e) What is the net entropy change of the ice + original water system as it reaches the equilibrium temperature?
GO An insulated Thermos contains 130 g of water at 80.0°C. You put in a 12.0 g ice cube at 0°C to form a system of ice + original water. (a) What is the equilibrium temperature of the system? What are the entropy changes of the water that was originally the ice cube (b) as it melts and (c) as it warms to the equilibrium temperature? (d) What is the entropy change of the original water as it cools to the equilibrium temperature? (e) What is the net entropy change of the ice + original water system as it reaches the equilibrium temperature?
GO An insulated Thermos contains 130 g of water at 80.0°C. You put in a 12.0 g ice cube at 0°C to form a system of ice + original water. (a) What is the equilibrium temperature of the system? What are the entropy changes of the water that was originally the ice cube (b) as it melts and (c) as it warms to the equilibrium temperature? (d) What is the entropy change of the original water as it cools to the equilibrium temperature? (e) What is the net entropy change of the ice + original water system as it reaches the equilibrium temperature?
1. The average KE and temperature in Kelvin of the molecules of a gas are related by the
equation KE = 3/2 KT where k is the Boltzmann constant 1.38 x 10 m² kg s².
The diagram shows the energy levels for a Hydrogen atom.
Energy/eV
0.00
-1.51
3.39
13.58
Use this information to show that Hydrogen at room temperature will not emit light.
2. When hydrogen burns in oxygen 241.8 kJ of energy are released per mole. Show that this
reaction can produce light.
3. By using the fact that around any closed loop the sum of the EMFS = the sum of the PDs. Write
equations for the two loops shown in the cct below.
40
ΔΩ
I₂
4V
(loop1
20 (loop2) 2v
I+12
Use these equations to show that the current flowing through the 20 resistor is 0.75A
5. A potential divider circuit is made by stretching a 1 m long wire with a resistance of 0.1 per cm
from A to B as shown.
8V
A
100cm
B
sliding contact
5Ω
A varying PD is achieved across the 5 Q resistor by moving the slider along the resistance wire.
Calculate the distance from A when the PD across the 5 Q resistor is 6 V.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
The Second Law of Thermodynamics: Heat Flow, Entropy, and Microstates; Author: Professor Dave Explains;https://www.youtube.com/watch?v=MrwW4w2nAMc;License: Standard YouTube License, CC-BY