If 100mL of boiling water is combined with 25g of ice, what will its

temperature be when equilibrium is reached?

You'll get brainlest but I need it rn rn

85ºC

Explanation:

Assume that all the heat released by the boiling water is absorbed by the ice.

1. Heat released by the 100 mL liquid boiling water

100 mL * 100g/mL = 100g Q = mass * specific heat * ΔT Q = 100g * 4.186J/g * (100ºC - T)

2. Heat absorbed by the 25 g of ice

a) Heat to melt the ice

Assume the initial temperature of ice is 0ºC Heat to melt the ice = latent fusion heat * mass L = 334J/g * 25g = 8,350J

b) Heat to increase of liquid water from 0ºC until the equilibrium temperature

Q = mass * specitif heat * ΔT Q = 25g * 4.186J/g * (T - 25ºC)

3. Set your final equation:

100g * 4.186J/g * (100ºC - T) = 25g * 4.186J/g * (T - 25ºC)

Solve:

41,860 - 418.6T = 104.25T - 2,616.25 523.25T = 44,476.25 T = 85ºC