Question

A 6.5g mass of ice with a temperature of 0 degree C is dropped Into Lake Erie, which is at a temperature of 15 degree C. What will be the final temperature, in Celsius, of the lake plus the ice? (You don't need to enter units. Just the numerical part of the answer.) If the latent heat of fusion of ice is 334kJ/kg and the specific heat of water is 4.18kJ kg-1 K-1 how much heat, in Joules, must 8.6g of ice absorb to reach the final temperature in part a? (No units necessary in your answer, just the numerical part of your answer.) Does the water originally in the lake lose or gain heat? Lose Gain How much heat is lost/gained by the lake as the ice cube melts? More heat than was calculated in part b. The same heat that was calculated in part b. Less heat than was calculated in part b. We do not have enough information.

Answer 1

Given

mass of the ice m = 6.5 g

temperature of lake water = 15⁰C = 288 K;

temperature of ice = 0⁰C = 273 K

specific heat capacity of water C = 4.18 J/g-K

the latent heat of fusion of ice L = 334 J/g

(1)

The final temperature will be 15⁰ C as the lake has too much mass as compared to the ice and thus it's temperature change will be extremely small to be noticed.

(2)

heat absorbed by ice to melt (Q₁) = m*L = 6.5*334 = 2171 J

heat absorbed by the melted ice (at 0⁰C) to reach 15⁰C (Q₂) = m*C*ΔT = 6.5*4.18*15 =  407.55 K

Hence the net heat absorbed by the ice ΔQ = Q₁ + Q₂ = 2578.55 J

(3)

The water originally in the lake loses water.

(4)

Zeroth law of thermodynamics which states that when two bodies are in thermal contact with each other then the heat lost by the body at higher temperature is equal to the heat gained by the body at lower temperature.

Hence the heat lost by the lake = ΔQ = 2578.55 J (the same heat that was calculated in part 3)