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 × 103 J kg-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 close to :

Question

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 × 103 J kg-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 close to : (A) 1.67 × 105 J (B) 1.68 × 106 J (C) 1.51 × 105 J (D) 1.71 × 107 J

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/c839ff060b5c3f99e7a5912164549c16-.png" /> heat required to freeze 5 kg water =5 × 336 × 103 =1680 × 103 Joule ⇒ Q 1=1680 KJ for carnot's cycle (Q2/Q1)=(T2/T1) ( Q 2/1680)=(300/273) Q 2=1680 × (300/273) KJ W=Q2-Q1 =1680((300/273)-1) =(1680 × 27/273) × 103 J =166.15 × 103 J =1.66 × 105 KJ

$1.67 \times 1 0^{5} J$

$1.68 \times 1 0^{6} J$

$1.51 \times 1 0^{5} J$

$1.71 \times 1 0^{7} J$

Q. 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 5kg of water at 0∘C is converted into ice at 0∘C by the freezer, then the energy consumed by the freezer is close to :

1.67×105J

1.68×106J

1.51×105J

1.71×107J

heat required to freeze 5kg water =5×336×103 =1680×103 Joule ⇒Q1​=1680KJ for carnot's cycle Q1​Q2​​=T1​T2​​ 1680Q2​​=273300​ Q2​=1680×273300​KJ W=Q2​−Q1​ =1680(273300​−1) =2731680×27​×103J =166.15×103J =1.66×105KJ

$1.67 \times 1 0^{5} J$

$1.68 \times 1 0^{6} J$

$1.51 \times 1 0^{5} J$

$1.71 \times 1 0^{7} J$