Question

A copper can with heat capacity 600 J/°C contains 200 g of water at 10°C. Dry steam at 100°C is passed in while the water is stirred and until the whole reaches a temperature of 30°C. Calculate the mass of steam condensed.

Assume water:
C value of 4200J/(kg°C)
L value of 2 300 000J/kg (vaporisation)

And copper:
C value of 300J/(kg°C)

Please show working.

Answer 1

Answer:

Let the mass of ice be m kg. Then mass of water is 0.2-m kg.

Heat loss by steam in converting to water and lowering temperature from 100

o

C to 50

o

C will be used to raise temperature of calorimeter and initial ice and water mi\timesture. First ice will convert to water and then the temperature will rise to 50

o

C

⟹(0.1)(0.42×10

3

)(50−0)+m(3.36×10

5

)+(0.2)(4.2×10

3

)(50−0)=(0.03)(22.5×10

5

)+ (0.03)(4.2×10

3

)(100−50)

⟹(0.18)(4.2×10

3

)(50)+m(3.36×10

5

)=(0.03)(22.5×10

5

)

⟹m≈88g

⟹

200−m

m

=1:1.26