Answer:
Explanation:
Given
Temperature of the gas is increased from 100 to 200
Also we know that average kinetic energy of the molecules is

Where
R=Gas constant
=Avogadro's number
T=Temperature in kelvin

So kinetic energy increases by


Answer:
The correct option is;
The graduate cylinder with more water has more thermal energy because it is holding more water molecules
Explanation:
Given that the thermal energy of the system is the energy possessed by the system by virtue of the increased motion of the particles by virtue of a transfer of heat, when the content of the system is heated
The thermal energy, Q is given by the following equation;
Q = Mass, m × The specific heat capacity, C × The change in temperature, ΔT
Given that the graduated cylinder with more water has more mass and therefore, more water molecules, than the cylinder with less water, the cylinder with more water has more thermal energy.
<h3>
Answer:</h3>
117.6 Joules
<h3>
Explanation:</h3>
<u>We are given;</u>
- Force of the dog is 24 N
- Distance upward is 4.9 m
We are required to calculate the work done
- Work done is the product of force and distance
- That is; Work done = Force × distance
- It is measured in Joules.
In this case;
Force applied is equivalent to the weight of the dog.
Work done = 24 N × 4.9 m
= 117.6 Joules
Hence, the work done in lifting the dog is 117.6 Joules
Answer: 361° C
Explanation:
Given
Initial pressure of the gas, P1 = 294 kPa
Final pressure of the gas, P2 = 500 kPa
Initial temperature of the gas, T1 = 100° C = 100 + 273 K = 373 K
Final temperature of the gas, T2 = ?
Let us assume that the gas is an ideal gas, then we use the equation below to solve
T2/T1 = P2/P1
T2 = T1 * (P2/P1)
T2 = (100 + 273) * (500 / 294)
T2 = 373 * (500 / 294)
T2 = 373 * 1.7
T2 = 634 K
T2 = 634 K - 273 K = 361° C
Answer:
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Explanation: