One mole of ice at 0°C is added to two moles of water at 50°C under a constant external pressure of 1 atm. This process is carried out in a thermally insulated vessel, such that no heat can escape the vessel. When equilibrium is reached, all of the ice has melted. Given: Δ H f u s i o n ∘ ( H 2 O a t 0 ∘ C ) = 6.01 k J m o l C p ( H 2 O , l i q u i d ) = 75 J m o l K a) (7 Points) Calculate the final temperature of the mixture (in °C). Show all work to support your answer.

T2 = 29.79°C

Explanation:

Equliibrium signifies that heat loss = heat gained

Heat gained by Ice;

H = ML

Mass, M = Number of moles * Molar mass = 1 * 18 = 18g

l = 6.01 k J m o l = 334 J/g

C = 4.186 J/g

H =  18(334)

H = 6012

Heat lost by water

H = MCΔT

H = 18 * 4.186 * (50 - T2)

H = 3767.4 - 75.348T2

Since H = H, we have;

6012 = 3767.4 - 75.348T2

- 75.348T2 = 3767 - 6012

T2 = 2245 / 75.348

T2 = 29.79°C

water is used to store them because it would keep the radiation from getting out in the open and causing harm to the workers