Question

In a container of negligible mass, add 0.140 kg of ice at -15.0 ° C to 0.190 kg of water at 35.0 ° C.

a) If no heat is lost to the environment, what final temperature does the system reach?

b) At the final temperature, how many kilograms are there of ice and how many kilograms of liquid water?

Answer 1

Given:

Let the mass of ice be M₁

Mass of ice, M₁ = 0.14 kg

Let the temperature of ice be T₁

Temperature of ice, T₁ = -15°C

Let the mass of water be M₂

Mass of water, M₂ = 0.19 kg

Temperature of water, T₂ = 35°C

Specific heat of a substanc is defined as the amount of heat required to raise the temperature of 1 g of a substance by 1°C and is Denoted by C.

When, the ice is added to the water, due to the temperature difference, there is a flow of heat from water to the ice.

It is important to understand the phase changes that may occur in order to write the mathematical expression.

Heat lost from Water = Heat gained by ice

Heat lost from water= M₂ x C_w X (T₂-T₃)

Heat gained by ice = M₁ x L + M₁ x C_ice x ( 0-T₁) + M₁ x C_w x ( T₃ - 0 )

L is the latent heat of fusion of ice which is the amount of energy associated with the conversion of 1 kg of ice into

1 kg of water. L = 3.33 x 10⁵

Therefore total energy to heat the ice from -15°C to 0° and the melt the ice fully:

here, (M x L) + M₁ x C_ice x (0 -(-15)) = 46,620 + (0.14 x 2090 x 15) = 51,009 J

Total energy that the water can transfer to the ice without freezing is when the water reaches 0°C.

H = 0.19 x 4186 x 35 = 27,836.9

Energy required to heat ice to 0°C = (0.15 x 2090 x 15) = 4,389 J

Energy required to melt 0.14 kg of ice at 0°C = 46,620 J

Total heat in water after heating ice to 0°C and when nothing of the ice has melted yet = 27,836.9 - 4,389 = 23,447.9 J

Let us calculate the mass of ice (say M) that will melt when supplied with 23,447.9 J :

23,447.9 = M x 3.33 x 10⁵

M = 0.07 kg

Therefore when the water reaches 0°C , 0.07 kg of the ice has melted. When the water reaches 0°C, the mixture is at thermal equilibrium as both the ice and water is at 0°C.

Therefore:

(a) Temperature of the mixture at thermal equilibrium: 0°C

(b) Mass of water at 0°C = Initial mass + (mass of melted ice) = 0.19 + 0.07 = 0.26 kg

Mass of ice = initial mass of ice - melted ice = 0.14 - 0.07 = 0.07 kg