JEE Advanced · Physics · 12. Thermal Properties

A current carrying wire heats a metal rod. The wire provides a constant power $(P)$ to the rod. The metal rod is enclosed in an insulated container. It is observed that the temperature $(T)$ in the metal rod changes with time $(t)$ as:
$T (t) = T_{0} (1 + β t^{1 / 4})$
where $β$ is a constant with appropriate dimension while $T_{0}$ is a constant with dimension of temperature. The heat capacity of the metal is:

A $\frac{4 P (T (t) - T_{0})}{β^{4} T_{0}^{2}}$
B $\frac{4 P {(T (t) - T_{0})}^{2}}{β^{4} T_{0}^{3}}$
C $\frac{4 P {(T (t) - T_{0})}^{4}}{β^{4} T_{0}^{5}}$
D $\frac{4 P {(T (t) - T_{0})}^{3}}{β^{4} T_{0}^{4}}$

Verified Solution

Answer & Solution

Correct Answer

(D) $\frac{4 P {(T (t) - T_{0})}^{3}}{β^{4} T_{0}^{4}}$

Step-by-step Solution

Detailed explanation

Heat required for small increment in temperature of a substance is given by

d Q = m . s . d T

where,

m =

mass of substances
or

s =

specific heat of substance
Now, heat capacity,

H = m . s .

∴ d Q = H . d T

\Rightarrow

Rate of heat absorption by the rod,

\frac{d Q}{d t} = P = H \cdot \frac{d T}{d t}

.......(1)
But given that, temperature of rod as a function of time is

T = T_{0} (1 + β t^{1 / 4})

......(2)

\Rightarrow \frac{d T}{d t} = T_{0} \cdot β \cdot \frac{1}{4} t^{- \frac{3}{4}}

Putting it in (1) gives

P = H \cdot T_{0} \cdot β \cdot \frac{1}{4} \cdot t^{- \frac{3}{4}}

\Rightarrow H = \frac{4 P \cdot t^{3 / 4}}{T_{0} \cdot β}

\Rightarrow H = 4 P \frac{{(T - T_{0})}^{3}}{T_{0}^{4} \cdot B^{4}}

Now using equation

(2)

T - T_{0} = T_{0} β \cdot t^{1 / 4}

\Rightarrow t^{3 / 4} = T_{0} {(\frac{T - T_{0}}{T_{0} β})}^{3}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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From JEE Advanced

Answer the following by appropriately matching the lists based on the information given in the paragraph.
In a thermodynamics process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by

T Δ X,

where

T

is temperature of the system and

Δ X

is the infinitesimal change in a thermodynamic quantity

X

of the system. For a mole of monatomic ideal gas

X = \frac{3}{2} R l n (\frac{T}{T_{A}}) + R l n (\frac{V}{V_{A}}) .

Here,

R

is gas constant,

V

is volume of gas,

T_{A}

and

V_{A}

are constants.
The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.

	Column - I		Column - II
(a)	Work done by the system in process $1 \to 2 \to 3$	(p)	$\frac{1}{3} R T_{0} l n 2$
(b)	Change in internal energy in process $1 \to 2 \to 3$	(q)	$\frac{1}{3} R T_{0}$
(c)	Heat absorbed by the system in process $1 \to 2 \to 3$	(r)	$R T_{0}$
(d)	Heat absorbed by the system in process $1 \to 2$	(s)	$\frac{4}{3} R T_{0}$
		(t)	$\frac{1}{3} R T_{0} (3 + l n 2)$
		(u)	$\frac{5}{6} R T_{0}$

If the process on one mole of monatomic ideal gas is as shown in the

P V

-diagram with

P_{0} V_{0} = \frac{1}{3} R T_{0},

the correct match is

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