Question

In a container of negligible mass, 0.400 kg of ice at an initial temperature of -29.0 ∘C is mixed with a mass m of water that has an initial temperature of 80.0∘C. No heat is lost to the surroundings.Part AIf the final temperature of the system is 26.0 ∘C, what is the mass m of the water that was initially at 80.0∘C?

Answer 1

Answer:

1 kg

Explanation:

The container has negligible mass and no heat is loss to the surrounding.

Mass of ice = 0.4kg, initial temperature of ice = -29oC, final temperature of the mixture = 26oC, mass of water (m2) = ?kg, initial temperature of water = 80oC, c ( specific heat capacity of water ) = 4200J/kg.K, Lf = heat of fusion of water = 3.36 × 10^5 J/kg

Using the formula:

Quantity of heat gain by ice = Quantity of heat loss by water

Quantity of heat gain by ice = mass of ice × heat of fusion of ice + mass of water × specific heat capacity of water = (0.4 × 3.36 × 10^ 5) + (0.4 × 4200 × (26- (-29) = 13.44 × 10^4 + 9.24 × 10^ 4 = 22.68 × 10^4 J

Quantity of heat loss by water = m2cΔT

Quantity of heat loss by water = m2 ×4200× (80 - 26) = m(226800)

since heat gain = heat loss

22.68 × 10^4 = 226800 m2

divide both side by 226800

226800 / 226800 = m2

m2 = 1 kg