A metal cylinder of mass 0.5 kg is heated electrically by a 12 W heater in a room at 15ºC. The cylinder temperature rises uniformly to 25ºC in 5 min and finally becomes constant at 45ºC. Assuming that the rate of heat loss is proportional to the excess temperature over the surroundings

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

The rate of loss of heat of the cylinder to surrounding at 20ºC is 2 W

The rate of loss of heat of the cylinder to surrounding at 45ºC is 2 W

Specific heat capacity of metal is $\frac{240}{\ln (3 / 2)} J / {kg}^{\circ} C$

None of these

answer is A, C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

In final state, $\frac{{dQ}_{e}}{dt} = k (45 - 15) = 12$

$\begin{array}{l} K = \frac{2}{5}; at 20^{\circ} C, \frac{{dQ}_{e}}{dt} = k (20 - 15) = \frac{2}{5} \times 5 = 2 watt. \\ m . s . \frac{dT}{dt} = 12 - \frac{2}{5} (T - 15) \Rightarrow m . s \cdot \frac{dT}{dt} = 18 - \frac{2}{5} T \\ m . s \cdot \int_{15}^{25} \frac{dT}{18 - \frac{2}{5} T} = \int_{0}^{300} dt \Rightarrow s = \frac{240}{\ln (\frac{3}{2})} J / kg -^{\circ} C \end{array}$

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET