Question

1 gram of ice is passed into a container having 2 grams of water at 35C the equilibrium temperature of the mixture is?​

Answer 1

Answer:

Explanation:

here's your answer

Answer 2

Given info : 1 gram of ice is passed into a container having 2 grams of water at 35°C .

To find : the equilibrium temperature of the mixture is ..

Solution : Heat needed by ice to change its state, H₁ = mLf , where Lf is latent heat of fusion of ice.

so, H₁ = 1 g × 80 cal/g = 80 cal

now heat lost by water to decrease its temperature to 0°C, H₂ = ms(T₂ - T₁)

= 2 g × 4.2 cal/g/°C × (35 - 0) = 292 cal

here you see, H₂ > H₁

It means ice completely melted and then it changed into water and was raised its temperature.

Let equilibrium temperature is T

Now, heat gained by ice = heat lost by water

⇒ H₁ + 1g × 4.2 cal/g/°C ×  T = 2g × 4.2 cal/g/°C × (35 - T)

⇒ 80 = 292 - 8.4T - 4.2 T

⇒ 212 = 12.6T

⇒ T = 16.8°C

Therefore the equilbrium temperature of the mixture is 16.8°C