If 1.40 g of steam at 100.0 ∘C condenses into 40.0 g of water initially at 25.0 ∘C, in an insulated container what is the final temperature?

Specific latent heat of vaporization of water = 2260 J/g
Specific heat capacity of water = 4.18 J/g°C

Let T°C be the final temperature of the mixture.

Heat released for 1.40 g of 100°C steam converted to 100°C water = (1.40 g) × (2260 J/g) …… [1]
Heat released for 1.40 g of 100°C water converted to T°C water = (1.40 g) × (4.18 J/g°C) × (100.0 - T)°C …… [2]
Heat absorbed for 40.0 g of 25°C water converted to T°C water = (40.0 g) × (4.18 J/g°C) × (T - 25.0)°C …… [3]

[1] + [2] = [3]
1.40 × 2260 + 1.40 × 4.18 + (100.0 - T) = 40.0 × 4.18 × (T - 25.0)
3164 + 585.2 - 5.852T = 167.2T - 4180
173.052T = 7929.2
T = 45.8

Final temperature = 45.8°C