Which of these is at thermal equilibrium? A- 25° 5° B- 25° 25° C- 5° 5° D- 5° 25°

A hot air balloon is filled with 1.45 × 10 6 L of an ideal gas on a cool morning ( 11 ∘ C ) . The air is heated to 109 ∘ C . Wha

just olya [345]

Answer:

The volume of air after the balloon is heated = 1.95 × 10⁶ L

Explanation:

Charles Law: Charles' law states that the volume of a given mass of gas is directly proportional to the temperature in Kelvin, provided that the pressure remains constant.

It can be expressed mathematically as,

V₁/T₁ = V₂/T₂

Making V₂ The subject of the equation

V₂ = (V₁/T₁)T₂.................... Equation 1

Where V₁ = Initial Volume, T₁ = Initial Temperature, V₂ = Final Volume, T₂ = final Temperature

Given: V₁ = 1.45 × 10⁶ L, T₁ = 11 °C = (11 + 273) K = 284 K, T₂ = 109 °C = (109 + 273) = 382 K.

Substituting these values into equation 1 above,

V₂ = (1.45×10⁶)382/284

V₂ = 1.95 × 10⁶ L

Therefore the volume of air after the balloon is heated = 1.95 × 10⁶ L

6 0

3 years ago

A particle of ink in a ink-jet printer carries a charge of -8x 10^-13 C and is deflected onto paper force of 3.2x10^-4. Find the

vitfil [10]

Beef and cheddar is your answer

3 0

3 years ago

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

Sliva [168]

To solve the problem it is necessary to take into account the concepts related to energy efficiency in the engines, the work done, the heat input in the systems, the exchange and loss of heat in the soupy the radius between the work done the lost heat ( efficiency).

By definition the efficiency of the heat engine is

$\epsilon = 1- \frac{T_c}{T_h}$

Where,

$T_c =$ Temperature at the room

$T_h =$ Temperature of the soup

The work done is defined as,

$dW = \epsilon(dQ_h)$

Where $Q_h$ represents the input heat and at the same time is defined as

$dQ_h =c_v (dT_h)$

Where,

$c_V =$ Specific Heat

The change at the work would be defined then as

$dW = \epsilon(dQ_h)$

$dW = \epsilon c_v (dT_h)$

$dW = (1-\frac{T_c}{T_h})c_v (dT_h)$

$W = \int dW = \int (1-\frac{T_c}{T_h})c_v (dT_h)$

$W = c_v (T_h-T_c)-c_v T_c ln(\frac{T_h}{T_c})$

On the other hand we have that the heat lost by the soup is equal to

$dQ_h =c_v (dT_h)$

$Q_h =c_v (T_h-T_c)$

The ratio between both would be,

$\frac{W}{Q_h} = \frac{c_v (T_h-T_c)-c_v T_c ln(\frac{T_h}{T_c})}{c_v (T_h-T_c)}$

$\frac{W}{Q_h} = \frac{1+ln(\frac{T_h}{T_c})}{1-\frac{T_h}{T_c}}$

Replacing with our values we have,

$\frac{W}{Q_h} = 1+\frac{ln(\frac{340K}{300K})}{1-\frac{340K}{300K}}$

$\frac{W}{Q_h} = 0.0613$

Therefore the fraction of heat lost by the soup that can be turned into useable work by the engine is 0.0613.

5 0

3 years ago

What does it mean when work is positive?

dmitriy555 [2]

Answer:

When force and displacement are in the same direction, the work performed on an object is said to be positive work. Example: When a body moves on the horizontal surface, force and displacement act in the forward path. The work is done in this case known as Positive work.

Explanation:

Hope this helps you

5 0

3 years ago

Read 2 more answers

A research study that produces a correlation coefficient of -0.90 indicates _

mario62 [17]

The absolute awser is b

8 0

3 years ago

Which of these is at thermal equilibrium?A- 25° 5°B- 25° 25°C- 5° 5°D- 5° 25°​

Which of these is at thermal equilibrium?A- 25° 5°B- 25° 25°C- 5° 5°D- 5° 25°