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Heat and Thermodynamics question

2024 · Shift 1 · Q42
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Heat and Thermodynamics question

2024 · Shift 1 · Q42

JEE AdvancedPhysicsHeat and ThermodynamicsNumerical+4 / −1
The specific heat capacity of a substance is temperature dependent and is given by the formula C=kTC=k TC=kT, where kkk is a constant of suitable dimensions in SI units, and TTT is the absolute temperature. If the heat required to raise the temperature of 1 kg1 \mathrm{~kg}1 kg of the substance from −73∘C-73^{\circ} \mathrm{C}−73∘C to 27∘C27^{\circ} \mathrm{C}27∘C is nkn knk, the value of nnn is ‾\underline{\hspace{2cm}}​. [Given: 0 K=−273∘C0 \mathrm{~K}=-273{ }^{\circ} \mathrm{C}0 K=−273∘C.]
Numerical answer
View written solutionFree

Correct answer: 25000

Step-by-step Solution:

  1. Identify Given Information:

    • Mass of the substance, m=1 kgm = 1 \mathrm{~kg}m=1 kg.
    • Temperature-dependent specific heat capacity, C=kTC = kTC=kT, where TTT is the absolute temperature in Kelvin.
    • Initial temperature, Tinitial(∘C)=−73∘CT_{initial}(^{\circ} \mathrm{C}) = -73^{\circ} \mathrm{C}Tinitial​(∘C)=−73∘C.
    • Final temperature, Tfinal(∘C)=27∘CT_{final}(^{\circ} \mathrm{C}) = 27^{\circ} \mathrm{C}Tfinal​(∘C)=27∘C.
    • Total heat required, Q=nkQ = nkQ=nk.
    • Conversion factor: T(K)=T(∘C)+273T(\mathrm{K}) = T(^{\circ} \mathrm{C}) + 273T(K)=T(∘C)+273.
  2. Convert Temperatures to Kelvin: The formula for specific heat capacity is given in terms of absolute temperature (Kelvin). Therefore, we must convert the given temperatures from Celsius to Kelvin.

    • Initial temperature, T1=−73+273=200 KT_1 = -73 + 273 = 200 \mathrm{~K}T1​=−73+273=200 K.
    • Final temperature, T2=27+273=300 KT_2 = 27 + 273 = 300 \mathrm{~K}T2​=27+273=300 K.
  3. Set up the Heat Calculation: When the specific heat capacity is a function of temperature, the heat required (QQQ) to raise the temperature of a mass (mmm) from T1T_1T1​ to T2T_2T2​ is given by the integral: Q=∫T1T2mC(T) dTQ = \int_{T_1}^{T_2} mC(T) \, dTQ=∫T1​T2​​mC(T)dT

  4. Substitute the Given Specific Heat Capacity: We are given C(T)=kTC(T) = kTC(T)=kT. Substituting this into the integral: Q=∫T1T2m(kT) dTQ = \int_{T_1}^{T_2} m(kT) \, dTQ=∫T1​T2​​m(kT)dT

  5. Evaluate the Integral: Since mmm and kkk are constants, we can take them out of the integral: Q=mk∫T1T2T dTQ = mk \int_{T_1}^{T_2} T \, dTQ=mk∫T1​T2​​TdT The integral of TTT with respect to TTT is T22\frac{T^2}{2}2T2​. Evaluating this from T1T_1T1​ to T2T_2T2​: Q=mk[T22]T1T2Q = mk \left[ \frac{T^2}{2} \right]_{T_1}^{T_2}Q=mk[2T2​]T1​T2​​ Q=mk(T222−T122)Q = mk \left( \frac{T_2^2}{2} - \frac{T_1^2}{2} \right)Q=mk(2T22​​−2T12​​) Q=mk2(T22−T12)Q = \frac{mk}{2} (T_2^2 - T_1^2)Q=2mk​(T22​−T12​)

  6. Substitute Numerical Values: Now, we plug in the values for mmm, T1T_1T1​, and T2T_2T2​:

    • m=1 kgm = 1 \mathrm{~kg}m=1 kg
    • T1=200 KT_1 = 200 \mathrm{~K}T1​=200 K
    • T2=300 KT_2 = 300 \mathrm{~K}T2​=300 K

    Q=(1)k2((300)2−(200)2)Q = \frac{(1)k}{2} ( (300)^2 - (200)^2 )Q=2(1)k​((300)2−(200)2) Q=k2(90000−40000)Q = \frac{k}{2} ( 90000 - 40000 )Q=2k​(90000−40000) Q=k2(50000)Q = \frac{k}{2} ( 50000 )Q=2k​(50000) Q=25000kQ = 25000 kQ=25000k

  7. Determine the Value of n: The problem states that the heat required is Q=nkQ = nkQ=nk. Comparing this with our calculated result: nk=25000knk = 25000 knk=25000k Dividing both sides by kkk, we get: n=25000n = 25000n=25000

Final Answer:

The value of nnn is 25000.

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