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 = 100gQ = mass × specific heat × ΔTQ = 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 0ºCHeat to melt the ice = latent fusion heat × massL = 334J/g × 25g = 8,350J

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

Q = mass × specitif heat × ΔTQ = 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 - 2,616.25523.25T = 44,476.25T = 85ºC

10 neutrons is the answer