Specific Heats MCQ Quiz in मल्याळम - Objective Question with Answer for Specific Heats - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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നേടുക Specific Heats ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Specific Heats MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Specific Heats MCQ Objective Questions

Top Specific Heats MCQ Objective Questions

Specific Heats Question 1:

Statement (I): The specific heat at constant pressure for an ideal gas is always greater than the specific heat at constant volume.

Statement (II): Heat added at constant volume is not utilized for doing any external work.

  1. Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I)
  2. Both Statement (I) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I)
  3. Statement (I) is true but Statement (II) is false
  4. Statement (I) is false but Statement (II) is true

Answer (Detailed Solution Below)

Option 1 : Both Statement (I) and Statement (II) are individually true and Statement (II) is the correct explanation of Statement (I)

Specific Heats Question 1 Detailed Solution

Concept:

In case of constant volume process the work done is zero, so the heat supplied is used to raise the internal energy of gas only.

As we know that, Q = U + W, where, Q = heat, U = Internal energy, W = work done

Therefore, Q = U...............(as W = 0)

In case of constant pressure process the heat is required to be supplied for expansion work in addition to increase in internal energy of the gas. So the heat supplied to raise the temperature of the gas by unit degree in case of constant pressure is more than the constant volume process.i.e. Q = U + W

It follows that CP is always greater than CV.

Alternate method:

CP  - CV  = R

C= CV  + R

So,  CP  >  CV

CP  = Specific heat at constant pressure kJ/kgK

CV = Specific heat at constant volume kJ/kgK

R = universal gas constant kJ/kgK

Specific Heats Question 2:

How much heat must be absorbed by ice of mass m  = 700 gm at −15° C to take it to the liquid state at 10° C ?

Given parameters. Specific heat of ice (15°C) = 2220 J/kg. K, water = 4187 J/kg.K , water = 4187 J/kg.K Heat of fusion of water = 333 KJ / kg 

  1. 233kJ Approx
  2. 30 kJ Approx
  3. 286 kJ Appro
  4. 256 kJ Approx 

Answer (Detailed Solution Below)

Option 3 : 286 kJ Appro

Specific Heats Question 2 Detailed Solution

Concept:

  • Heat required to raise the temperature of ice from -15°C to 0°C: Q₁ = m × c₁ × ΔT
  • Heat required to melt ice at 0°C: Q₂ = m × L
  • Heat required to raise the temperature of resulting water from 0°C to 10°C: Q₃ = m × c₂ × ΔT
  • Total heat required: Q = Q₁ + Q₂ + Q₃

 

Calculation:

Mass of ice, m = 700 g = 0.7 kg

Initial temperature of ice, T₁ = -15°C

Final temperature of water, T₂ = 10°C

Specific heat of ice, c₁ = 2220 J/kg·K

Specific heat of water, c₂ = 4187 J/kg·K

Latent heat of fusion of ice, L = 333 kJ/kg

⇒ Heat to raise ice temperature:

Q₁ = 0.7 × 2220 × (0 - (-15))

⇒ Q₁ = 0.7 × 2220 × 15

⇒ Q₁ = 23.31 kJ

 

⇒ Heat to melt ice:

Q₂ = 0.7 × 333

⇒ Q₂ = 233.1 kJ

 

⇒ Heat to raise water temperature:

Q₃ = 0.7 × 4187 × (10 - 0)

⇒ Q₃ = 0.7 × 4187 × 10

⇒ Q₃ = 29.3 kJ

 

⇒ Total heat required:

Q = Q₁ + Q₂ + Q₃

⇒ Q = 23.31 + 233.1 + 29.3

⇒ Q ≈ 286 kJ

∴ The total heat required is approximately 286 kJ.

Specific Heats Question 3:

A container filled with 2 kg of O2 is heated at constant pressure from 27°C to 127°C. The heat supplied in this process is:

  1. 205 kJ
  2. 100 kJ
  3. 225 kJ
  4. 182 kJ

Answer (Detailed Solution Below)

Option 4 : 182 kJ

Specific Heats Question 3 Detailed Solution

Concept:

Heat supplied at constant pressure = mCp(T2 - T1)

Calculation:

Given:

m = 2 kg, T1 = 27°C, T2 = 127°C, Cp of oxygen = 0.918 kJ/kg-K (Experimental results)

Heat supplied at constant pressure = mCp(T2 - T1)

= 2 × 0.918 × (127 - 27) = 183.6 kJ

Specific Heats Question 4:

The product of mass and specific heat is__________.

  1. latent heat
  2. difference in temperature 
  3. heat capacity
  4. heat transfer rate 

Answer (Detailed Solution Below)

Option 3 : heat capacity

Specific Heats Question 4 Detailed Solution

Explanation:

Heat Capacity (C):

It is the amount of energy or heat required to raise the temperature of a substance by 1 °C or 1 K.

It is the product of heat capacity (mc) and change in temperature (ΔT). It is given as -

Q = mcΔT = CΔT   ...... eq (1)

Q = heat required

C = mc = heat capacity of the substance (unit J/K or J/°C)

m = mass of the substance

c = heat capacity per unit mass

Specific heat (c):

It is the amount of energy or heat required to raise the temperature of the unit mass of a substance by 1 °C or 1 K. It is given by-

\(c =\frac{Q}{mΔ T}\)     eq.... (2)

Its unit is J/kg-K.

From eq(1) and (2)

C = mc ⇒ mass of substance × specific heat.

Specific Heats Question 5:

For a perfect gas the specific heat at constant pressure Cp and that at constant volume Cv are related as:-

  1. Cp = Cv
  2. Cp = Cv + R
  3. Cp = Cv - R
  4. Cp = Cv + log R

Answer (Detailed Solution Below)

Option 2 : Cp = Cv + R

Specific Heats Question 5 Detailed Solution

Concept:

  • Perfect gas or ideal gas is the hypothatical condition which obeys the ideal gas law or the real nature of gas that is why its ideal condition.
  • Specific heat is the quantity of heat required to raise the temperature of unit mass of the gas by 1 degree.
  • Specific heat at constant pressure CP is the quanity of heat required to raise 1 degree at constant pressure, and Specific heat CV is the quantity of heat required to raise 1 degree at constant volume.

the specific heat is at constant pressure (Cp) is greater then the specific heat at constant volume (CV) And the gas constant can be expressed as C- CV = R, this relation is termed as Mayer’s Formula.

Where,

R = Gas constant

n = molar mass of the substance

Cp= molar specific heat at constant pressure

C= molar specific heat at constant Volume

Specific Heats Question 6:

Which of the following options are correct? (where γ = ratio of specific heats, Cp = specific heat at constant pressure, Cv = specific heat at constant volume).

  1. \(\gamma = \frac{1}{{1 - \left( {\frac{{{C_v}}}{R}} \right)}}\)
  2. \(\gamma = \frac{1}{{1 - \left( {\frac{R}{{{C_p}}}} \right)}}\)
  3. \(\gamma = \frac{1}{{1 - \left( {\frac{{{C_p}}}{R}} \right)}}\)
  4. \(\gamma = \frac{1}{{1 - \left( {\frac{R}{{{C_v}}}} \right)}}\)

Answer (Detailed Solution Below)

Option 2 : \(\gamma = \frac{1}{{1 - \left( {\frac{R}{{{C_p}}}} \right)}}\)

Specific Heats Question 6 Detailed Solution

Explanation:

We know that

\(\gamma = \frac{{{C_p}}}{{{C_v}}}\)

By rearranging the above equation we get,

\(\frac{{{C_p}}}{{{C_v}}} = \frac{1}{{\frac{{{C_p} - {C_p} + {C_v}}}{{{C_p}}}}}\)

∴ \(\frac{{{C_p}}}{{{C_v}}} = \frac{1}{{1 - \left( {\frac{{{C_p} - {C_v}}}{{{C_p}}}} \right)}}\)

∴ \(\gamma =\frac{{{C_p}}}{{{C_v}}} = \frac{1}{{1 - \left( {\frac{R}{{{C_p}}}} \right)}}\)

Specific Heats Question 7:

Gases have

  1. Two specific heats
  2. Three specific heats
  3. One specific heat
  4. None of these

Answer (Detailed Solution Below)

Option 1 : Two specific heats

Specific Heats Question 7 Detailed Solution

The specific heats of gases are generally expressed as molar specific heats. Two specific heats are defined for gases, one for constant volume (CV) and one for constant pressure (CP). These two specific heats are related by gas constant(R) as followed

C- CV = R

Specific Heats Question 8:

Sl unit of specific heat capacity is 

  1. J/mk
  2. J/kgK
  3. J/kg
  4. J/K

Answer (Detailed Solution Below)

Option 2 : J/kgK

Specific Heats Question 8 Detailed Solution

Explanation:

Specific heat capacity:

  • Specific heat capacity (s) of a substance is defined as the amount of heat (​ΔQ) per unit mass of the substance that is required to raise the temperature (ΔT) by 1°C.

s = \(\frac{1}{m} \frac{ΔQ}{ΔT}\)

  • The unit of specific heat capacity is J/kg.K (J kg-1 K-1).

Specific Heats Question 9:

The value of specific heat for a gas undergoing polytropic process in terms of cv, polytropic index n and γ is:

  1. \({{c}_{v}}\)
  2. \({{c}_{v}}\frac{n-\gamma }{1-n}\)
  3. 0
  4. \({{c}_{v}}\frac{\gamma -n}{1-n}\)

Answer (Detailed Solution Below)

Option 4 : \({{c}_{v}}\frac{\gamma -n}{1-n}\)

Specific Heats Question 9 Detailed Solution

Explanation:

For polytropic process:

\({{W}_{12}}=\frac{{{P}_{1}}{{V}_{1}}-{{P}_{2}}{{V}_{2}}}{n-1}\)

\({{Q}_{12}}={{W}_{12}}+\text{ }\!\!\Delta\!\!\text{ }U=\frac{{{P}_{1}}{{V}_{1}}-{{P}_{2}}{{V}_{2}}}{n-1}+m{{c}_{v}}\left( {{T}_{2}}-{{T}_{1}} \right)=\frac{mR\left( {{T}_{1}}-{{T}_{2}} \right)}{n-1}+m{{c}_{v}}\left( {{T}_{2}}-{{T}_{1}} \right)\)

\({{Q}_{12}}=m\left( {{T}_{2}}-{{T}_{1}} \right)\left[ {{c}_{v}}-\frac{R}{n-1}~ \right]\)

\({{c}_{v}}-\frac{R}{n-1}={{c}_{v}}-\frac{{{c}_{p}}-{{c}_{v}}}{n-1}={{c}_{v}}-\frac{{{c}_{v}}\left( \gamma -1 \right)}{n-1}={{c}_{v}}\left[ 1-\frac{\left( \gamma -1 \right)}{n-1} \right]=\frac{{{c}_{v}}\left( n-\gamma \right)}{n-1}=\frac{{{c}_{v}}\left( \gamma -n \right)}{1-n}\)

\({{Q}_{12}}=m\left( {{T}_{2}}-{{T}_{1}} \right)\left[ {{c}_{v}}-\frac{R}{n-1}~ \right]=m\left( {{T}_{2}}-{{T}_{1}} \right)\frac{{{c}_{v}}\left( \gamma -n \right)}{1-n}\)

\({{Q}_{12}}=m\frac{{{c}_{v}}\left( \gamma -n \right)}{1-n}\left( {{T}_{2}}-{{T}_{1}} \right)=m{{c}_{n}}\left( {{T}_{2}}-{{T}_{1}} \right)\)

The polytropic specific heat of a gas is given by:

\({{c}_{n}}={{c}_{v}}\frac{\gamma -n}{1-n}\)

Specific Heats Question 10:

40 g of water at 60°C is poured into a vessel containing 50 g of water at 20°C. The final temperature recorded is 30°C. The thermal capacity of the vessel is _______ J/°C. Take specific heat capacity of water as 4.2 J/g°C.

Answer (Detailed Solution Below) 293.8 - 294.2

Specific Heats Question 10 Detailed Solution

06.08.2018.001410122

Heat lost by hot water = Heat gained by cold water + Heat gained by vessel

mh C ΔTh = mc C ΔTc + mvcv ΔTv

40 × 4.2 × (60 – 30) = 50 × 4.2 × (30 – 20) + mvcv (30 – 20)

Cv = mv cv = 294 J/°C.

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