Question

Steam at 100°C is condensed into a 54.0 g aluminum calorimeter cup containing 260 g of water at 25.0°C. Determine the amount of steam (in g) needed for the system to reach a final temperature of 64.0°C. The specific heat of aluminum is 900 J/(kg · °C).

Answer 1

Answer would be,

18.4486 g

Solution:-

Mass of calorimeter cup Mc=54g= 54 / 1000kg = 0.054kg, mass of water in the calorimeter Mw =260g = 260/1000 kg = 0.26kg,

initial temperature of the calorimeter and water = 25° C, final temperature expected = 64°C

We have the formula as,

Heat gain = heat loss ....... (assuming no heat is loss to the surrounding

Heat loss by steam = mLv + mcΔT

(heat loss in cooling the water to 64°C)

where m is the mass of steam in kg,

c is the specific heat capacity of water (4200J/(Kg.oC) and

Latent heat of vaporization of water = 22.6 × 10^5J/Kg is the heat loss in condensing water back to steam

Taking m out as common and on substitution of the values into the above formula we will get,

Heat loss by steam = m (Lv + cΔT) = m (( -22.6 × 10^5) + ( 4200 × (64-100)))

Heat loss by steam = m ( -2260000 - 151200)

Heat loss by steam = m ( - 2411200)

where the - sign signify loss

Heat gain by calorimeter and water inside = MccΔT + MwcΔT

On substitution of the values inside the formula we will get,

Heat gain by calorimeter and water = (0.054 ×900×(64-25)) + (0.26 ×4200×(64-25)

Heat gain by calorimeter and water = 1895.4 + 42588 = 44483.4 J

As we know Heat gain equals heat loss so that,

44483.4 = m (2411200)

On dividing both side by 2411200 we will get,

m = 44483.4 / 2411200 = 0.0184486 kg = 18.44865 g

Which is our required amount of steam needed for the system.