Question

Steam is passed into 22g of water at 20c . The mass of water that will be present when the water acquires a temp of 90c?

Answer 1

Heat lost by steam = heat gained by water.

Heat required to condense steam at 100°C to water at 100°C

Q1= mL = m×540

Heat required to lower temperature of water from 100°C to 90°C

Q2 = ms( delta temp)

= m×1×(100-90)°C

Heat lost by steam = Q1 + Q2

Now, heat gained by water to raise it from 20°C to 90°C

=ms(delta temp)

=22×1×(90-20)°C

=22×70

=1540.

Now,

1540= Q1+ Q2

1540= m(10+540)

550m = 1540

m= 1540÷550

=2.83 g

Thus total amount of water

= 22 + 2.83 g

= 24.83g

This is the required answer.

Answer 2

Explanation:

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