मराठी

What Amount of Heat Must Be Supplied to 2.0 X 10-2 Kg of Nitrogen (At Room Temperature) to Raise Its Temperature by 45 °C at Constant Pressure? (Molecular Mass of N2 = 28; R = 8.3 J Mol-1 K-1.) - Physics

Advertisements
Advertisements

प्रश्न

What amount of heat must be supplied to 2.0 x 10-2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol-1 K-1.)

उत्तर १

Mass of nitrogen, m = 2.0 × 10–2 kg = 20 g

Rise in temperature, ΔT = 45°C

Molecular mass of N2M = 28

Universal gas constant, R = 8.3 J mol–1 K–1

Number of moles, `n = m/M`

`= (2.0xx10^(-2)xx10^3)/28 = 0.714`

Molar specific heat at constant pressure for nitrogen, `C_P = 7/2 R`

`= 7/2 xx 8.3`

`= 29.05 J mol^(-1) K^(-1)`

The total amount of heat to be supplied is given by the relation:

ΔQ = nCΔT

= 0.714 × 29.05 × 45

= 933.38 J

Therefore, the amount of heat to be supplied is 933.38 J.

shaalaa.com

उत्तर २

Here, mass of gas,` m= 2xx 10^-2 kg` = 20 g

rise in temperature, `triangleT = 45 ^@C`

Heat required, triangleQ = ?, Molecular mass, M = 28

Number of moles , `n = m/M = 20/28 = 0.714`

As nitrogen is a diatomatic gas,molar specific heat at constant pressure is

`C_p = 7/2R = 7/2 xx 8.3 J mol^(-1) K^(-1)`

As `triangleQ = nC_p triangleT`

`:. triangleQ = 0.714 xx 7/2 xx 8.3 xx 45 J = 933.4 J`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 12: Thermodynamics - Exercises [पृष्ठ ३१६]

APPEARS IN

एनसीईआरटी Physics [English] Class 11
पाठ 12 Thermodynamics
Exercises | Q 2 | पृष्ठ ३१६

संबंधित प्रश्‍न

You have a choice of three metals A, B, and C, of specific heat capacities 900 Jkg-1 °C-1, 380 Jkg-1 °C-1 and 460 Jkg-1 °C-1 respectively, to make a calorimeter. Which material will you select? Justify your answer.


Water in lakes and ponds do not freeze at once in cold countries. Give a reason is support of your answer.


50 g of metal piece at 27°C requires 2400 J of heat energy so as to attain a temperature of 327°C . Calculate the specific heat capacity of the metal.


The specific heat capacity of water is :


Which principle is used to measure the specific heat capacity of a substance?


What is the specific heat capacity of boiling water?


The specific heat capacity of a body depends on _____________ .


Why do the farmers fill their fields with water on a cold winter night?

(b) 2000 J of heat energy is required to raise the temperature of 4 kg of a
metal by 3°c. Which expression gives the specific heat capacity of the metal?


If 10125 J of heat energy boils off 4.5 g of water at 100°C  to steam at 100°C, find the specific latent heat of steam.


Water boils at 120 °C in a pressure cooker. Explain the reason


Why is specific heat capacity taken as a measure of thermal inertia?


Explain, Why is it advisabile to pour cold water over burns, caused on human body, by hot solids?


Some heat is provided to a body to raise its temperature by 25°C. What will be the corresponding rise in temperature of the body as shown on the Kelvin scale?


The temperature of a lead piece of mass 400 g rises from 20°C to 50°C when 1560 J of heat is supplied to it. Calculate Specific heat capacity of lead.


The value of 'γ' for a gas is given as `gamma = 1 + 2/"f"`, where 'f ' is the number of degrees of freedom of freedom of a molecule of a gas. What is the ratio of `gamma_"monoatonic"//gamma_"diatomic"`?
Diatomic gas consists of rigid gas molecules


Water has the lowest specific heat capacity.


Match the columns:

Column ‘A’ Column ‘B’
The SI unit of specific heat capacity (a) Jkg−1°C−1
(b) kg/m3
(c) calorie

The molar specific heats of an ideal gas at constant pressure and volume are denoted by Cp and Cv, respectively. If `gamma = "C"_"p"/"C"_"v"` and R is the universal gas constant, then Cv is equal to ______.


When two kilocalories of heat are supplied to a system, the internal energy of the system increases by 5030 J and the work done by the gas against the external pressure is 3350 J. Calculate J, the mechanical equivalent of heat.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×