Question

2 kq kg of ice at -4 degress Celsius is heated at Water at room temperature of 25 degrees Celsius .how much heat is absorbed ​

Answer 1

893.8 [kJ].

Step-by-step explanation:

1) the total heat can be calculated as

E=Q₁+Q₂+Q₃, where Q₁ - energy to heat the ice from -4 to 0; Q₂ - energy to melt the ice; Q₃ - to heat the water at 0 °C to 25 °C;

2) Q₁=λ*m*Δt₁, ⇒ Q₁=2100*2*4=16800 [J];

3) Q₂=c₂*m; ⇒ Q₂=333500*2=667000 [J];

3) Q₃=c₃*m*Δt₂; ⇒ Q₃=4200*2*25=210000 [j];

4) finally, the required heat is

E=16800+667000+210000=893800 [J]=893.8 [kJ].