Respuesta :
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
