Find the amount of heat required to change steam that is at 120 degrees celcuis to ice at 20 degrees celcuis. Can someone give me the formula to solve this problem?
2007-12-03 15:51:57 · 3 answers · asked by Momo 5 in Science & Mathematics ➔ Physics
I'm sorry, the temperature was negative 20 degrees celcius!
The mass is 120g, thank you for the answers so far :D
2007-12-03 16:14:46 · update #1
We need some physical data first:
Specific heat of steam: .48 cal / g-oC
Specific heat of water: 1.00 cal / g-oC
Specific heat of ice: .50 cal / g-oC
Latent Heat of water ( Boiling ) : 540 cal / g
Latent Heat of water ( Freezing ): 80 cal / g
The heat Q required to change the temperature of something with mass m and specific heat c is
Q = m c ΔT
The heat Q required to change the state of a material with mass m and latent heat L is
Q = mL
We start with steam at 120 oC. The steps are:
Cool the steam from 120 to 100 oC.
Condense the steam into water at 100 oC.
Cool the water from 100 to 0 oC.
Freeze the water into ice at 0 oC.
Cool the ice to (I'm guessing) -20 oC.
For each gram of steam you start with, the energy required is then
( .48 cal / g-oC )(100 - 120 ) - 540 cal / g + ( 1.00 cal / g-oC )( 0 - 100 ) - 80 cal / g + ( .50 cal / g-oC ) ( -20 - 0 )
Note that each term is negative since you're removing heat from the system at each stage.
2007-12-03 16:08:00 · answer #1 · answered by jgoulden 7 · 1⤊ 0⤋
You have not stated the pressure. Ice cannot exist at 20 degrees celcius under atmospheric pressure. So please specify your problem perfectly.
If I assume that the pressure is such that the melting point of ice is 20 degrees celcius, then it can be worked out as under.
Find the boiling point of water corresponding to the pressure. Suppose it is 110 degrees celcius, it would mean that steam is supernheated. So first it has to cool from 120 to 110 degrees celcius. 1 g of steam will need 5 cal. of heat to cool by 10 degrees, 540 cal to condense to water, 80 cal to become water at 20 degrees and 80 cal to become ice totalling 705 cal per gram.
2007-12-03 16:07:57 · answer #2 · answered by Madhukar 7 · 0⤊ 0⤋
First of all at 20 degrees celcius, water is in liquid form and NOT solid form (ice).... or maybe you mean -20 degrees celcius?....Secondly, to solve this problem completely the mass of steam or liquid water MUST be known
Q = mc(t2-t1) + mL + mc(t4-t1)
= m(0.48)(120-100) + m 539) + m (1)(100-20)
2007-12-03 16:03:24 · answer #3 · answered by jamesyoy02 6 · 0⤊ 0⤋