A Carnot freezer takes heat from water at 0°C inside it and rejects it to the room at a temperature of 27°C. The latent heat of ice is 336×103Jkg−1. If 5 kg of water at 0°C is converted into ice at 0°C by the freezer, then the energy consumed by the freezer is closed to

By conservation of energyQ1+Wext=Q2⇒Wext=Q2−Q1For a Carnot freezer we useQ1Q2=T1T2....(1)Here Q1→ heat released by freezer in freezing ice Here Q1=mL=5×336×103(J/kg)×kgWe use by equation - (1)Q2=(T2T1)×Q1=300273×5×336×103Energy consumed by freezer in W=Q2−Q1⇒W=5×336×103(300273−1)⇒W=1.67×105J

A Carnot freezer takes heat from water at 0°C inside it and rejects it to the room at a temperature of 27°C. The latent heat of ice is 336×103Jkg−1. If 5 kg of water at 0°C is converted into ice at 0°C by the freezer, then the energy consumed by the freezer is closed to

By conservation of energyQ1+Wext=Q2⇒Wext=Q2−Q1For a Carnot freezer we useQ1Q2=T1T2....(1)Here Q1→ heat released by freezer in freezing ice Here Q1=mL=5×336×103(J/kg)×kgWe use by equation - (1)Q2=(T2T1)×Q1=300273×5×336×103Energy consumed by freezer in W=Q2−Q1⇒W=5×336×103(300273−1)⇒W=1.67×105J

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A Carnot freezer takes heat from water at $0 ° C$ inside it and rejects it to the room at a temperature of $27 ° C .$ The latent heat of ice is $336 \times 10^{3} J k g^{- 1} .$ If 5 kg of water at $0 ° C$ is converted into ice at $0 ° C$ by the freezer, then the energy consumed by the freezer is closed to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$1.67 \times 10^{5} J$

$1.51 \times 10^{5} J$

$1.68 \times 10^{6} J$

$1.71 \times 10^{7} J$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Question Image

By conservation of energy
$Q_{1} + W_{e x t} = Q_{2}$
$\Rightarrow W_{e x t} = Q_{2} - Q_{1}$
For a Carnot freezer we use
$\frac{Q_{1}}{Q_{2}} = \frac{T_{1}}{T_{2}} .... (1)$
Here $Q_{1} \to$ heat released by freezer in freezing ice
Here $Q_{1} = m L = 5 \times 336 \times 10^{3} (J / k g) \times k g$
We use by equation - (1)
$Q_{2} = (\frac{T_{2}}{T_{1}}) \times Q_{1} = \frac{300}{273} \times 5 \times 336 \times 10^{3}$
Energy consumed by freezer in
$W = Q_{2} - Q_{1}$
$\Rightarrow W = 5 \times 336 \times 10^{3} (\frac{300}{273} - 1)$
$\Rightarrow W = 1.67 \times 10^{5} J$