An insulated container is divided into two halves by a conducting wall. One half carries 600 g of water while other half carries ice. The rate of heat transfer from water to ice is constant $P$ , until thermal equilibrium is achieved. The temperature variation of water and ice with time is shown graphically, then (Specific heat of water is $4.2 J / g /^{0} C$ , latent heat of fusion of ice is 80 cal/ g ) (Choose incorrect statement)

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

The initial mass of ice in the container is 2400g

The initial mass of ice in the container is 1200g

The mass of water converted to ice till thermal equilibration is achieved is 150g

The mass of water, converted to ice till thermal equilibration is achieved is 200g

answer is B, D.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$P = \frac{d Q}{d t} = m_{1} s_{1} \frac{d θ_{1}}{d t} = 600 g \times 4.2 j / g^{0} C \times \frac{40^{0} C}{40 \times 60 s} = 42 W$

$N o w, m_{1} s_{1} \frac{d θ_{1}}{d t} = m_{2} s_{2} \frac{d θ_{2}}{d t} \Rightarrow m_{2} = 2400 g$

It takes (60 – 40) = 20 min to reach thermal equilibrium

$s o, P t = m . L \Rightarrow m = \frac{P t}{L} = \frac{42 \times 1200 J}{81 \times 4.2 J / g m} = 150 g$

$L = \frac{Q}{m} \Rightarrow c a l / g, C = m s \Rightarrow c a l /^{0} C, W = \frac{m s}{S_{w}} \Rightarrow g$

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

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

Take Free Test