Heat loss from M grams steam cooling and condensing to 50 C liquid = heat gained by 250 g water increasing from 22 to 50 C PLUS heat gained by beacker. Write that as an equation and solve for M.
You will need to know the specific heat of steam in the gas phase (about 0.4 cal/g C), as well as the heat of condensation (540 cal/g). You will also need the specific heat of the glass beaker. I will leave the latter up to you to find out.
M*[25*0.4 + 540 + 50*1.0] = (heat gained by original water and beaker)
Solve for M
What mass of steam initially at 125 degree celcius is needed to warm 250g of water in 100g beaker from 22 degrees celcius to 50 degrees celcius?
1 answer
1 answer