Steam at 100°C is condensed into a 38.0 g steel calorimeter cup containing 220 g of water at 29.0°C. Determine the amount of steam (in g) needed for the system to reach a final temperature of 48.0°C. The specific heat of steel is 490 J/(kg * °C).

Mass of calorimeter cup Mc = 38 g = 38 / 1000kg = 0.038 kg, mass of water in the calorimeter Mw =220g = 220/1000 kg = 0.22kg, initial temperature of the calorimeter and water = 29 oC, final temperature expected = 48oC

Using the formula

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 48oC) 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

Take m out as common and substitute the values into the above formula

Heat loss by steam = m (Lv + cΔT) = m (( -22.6 × 10^5) + ( 4200 × (48-100)))

Heat loss by steam = m ( -2260000 - 218400)

Heat loss by steam = m ( - 2478400) where the - sign signify loss

Heat gain by calorimeter and water inside = MccΔT + MwcΔT

substitute the values inside the formula

Heat gain by calorimeter and water = (0.038 ×490×(48-29)) + (0.22 ×4200×(48-29)

Heat gain by calorimeter and water = 353.7 + 17556 = 17909.78 J

Remember Heat gain equals heat loss

17909.78 = m (2478400)

divide both side by 2478400

m = 17909.78 / 2478400 = 0.0072 kg = 7.2 g

okpalawalter8 okpalawalter8 · Answer 2 · 2022-04-12T10:59:13+02:00

The amount of steam (in g) needed for the system to reach a final temperature of 48.0°C is mathematically given as

m= 7.2 g

What is the amount of steam (in g) needed for the system to reach a final temperature of 48.0°C.?

Question Parameter(s):

Steam at 100°C is condensed into a 38.0 g steel calorimeter cup containing 220 g of water at 29.0°C.

Generally, the equation for the Heat loss by steam is mathematically given as

qH= mLv + mcΔT

Therefore, Heat loss by steam

qH= m (Lv + cΔT)

qH = m (( -22.6 × 10^5) + ( 4200 × (48-100)))

qH=m ( - 2478400)

Heat gain

qH'=MccΔT + MwcΔT

qH'= = (0.038 ×490×(48-29)) + (0.22 ×4200×(48-29)

qH= 17909.78 J

In conclusion, Heat gain equals heat loss

m = 17909.78 / 2478400

m = 0.0072 kg

m= 7.2 g