An ice cube of dimensions 60 cm × 50 cm × 20 cm is placed in an insulation box of wall thickness 1 cm. The box keeping the ice cube at 0°C of temperature is brought to a room of temperature 40°C. The rate of melting of ice is approximately:
(Latent heat of fusion of ice is 3.4 × 10⁵ J kg^–1 and thermal conducting of insulation wall is 0.05 Wm^–1 ºC^–1 )

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

61×10^–5 kg s^–1

30×10^–5 kg s^–1

61×10^–1 kg s^–1

208 kg s^–1

answer is B.

(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

$\begin{array}{l} \frac{dQ}{dt} = \frac{KAΔT}{ℓ} \\ A = 2 (0 .6 \times 0 .5 + 0 .5 \times 0 .2 + 0 .2 \times 0 .6) \end{array}$

$\begin{array}{l} = 2 (0 .3 + 0 .1 + 0 .12) \\ = 2 (0 .4 + 0 .12) \\ = 2 (0 .52) \\ = 1 .04 m^{2} \\ R_{th} = \frac{ℓ}{KA} \Rightarrow \frac{1 \times 10^{- 2}}{0 .05 \times 1 .04} \Rightarrow \frac{10^{- 2}}{0 .052} \\ \frac{dQ}{dt} = \frac{ΔT}{R_{th}} \Rightarrow \frac{40 \times 0 .052}{10^{- 2}} \Rightarrow 2 .08 \times 10^{2} J / s \\ 2 .08 \times 10^{2} = m \times 3 .4 \times 10^{5} \\ m = \frac{2 .08}{3 .4 \times 10^{3}} \Rightarrow 0 .61 \times 10^{- 3} kg / s \\ = 61 \times 10^{- 5} Kg / s \end{array}$

Watch 3-min video & get full concept clarity

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

How to Join IIT after 10th

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

How to Join IIT after 10th