# Finding the Value of M in a Physics Homework Problem

• Julieh
In summary: The only problem with Q1 and Q2 individually is the erroneous 3030x106 in Q1 that @BvU mentioned.The other problem is what you did with these two expressions. Q1 is the heat gained by the ice and water mix, while Q2 is the heat lost by the steam. What will be the relationship between those two quantities?Q1 and Q2 will be in equilibrium, and the sum of their expressions will be zero. This means that the final mixture will have 0.300 kg of water in it.
Julieh

## Homework Statement

Suppose that 0.250 kg of ice and 0.300 kg of water in thermal equilibrium are heated to 60°C by being mixed with M kg of steam at 100°C. Use Lice melting = 3.30 x 105 J/kg, Lwater vaporizing= 2.30 x 106 J/kg, and cwater = 4.2 x 103 J/kg°C.
(a) Write an expression for Q1, the heat gained by the ice-and-water system. Be sure to include the heat involved in both melting the ice and raising the temperature of the melted ice and water
Write an expression for Q2, the heat lost by the steam. Include both the heat to condense the steam and to lower the temperature of the resulting liquid water. (3)
(c) Equate the sum of the expressions from (a) and (b) with zero, and find the value of M. How much water will be in the final mixture

## Homework Equations

heat gained=heat lost
q1=Mice*Lice+Cwater*Mice*change in temperature
q2=Msteam*Lvaporization+Cwater*Mwater*Change in themperature of water

## The Attempt at a Solution

when i sum up the equations together and equate them to 0 the answer that i got is -0.1 kg, and mass can not be negative

Hope you'll find the sub- and superscript buttons soon -- they look like x2 and x2 to the right of the camera icon

Please show your work (the equations that yield a negative mass) and we'll take it from there

BvU said:

Hope you'll find the sub- and superscript buttons soon -- they look like x2 and x2 to the right of the camera icon

Please show your work (the equations that yield a negative mass) and we'll take it from there
Q1+Q2=0
and then Q1 is :[0.250kg*(3030*10^6)+(4.2*10^3)*0.250*60] and i added Q2:[(M*2300000)+(4200*0.300*40)]
and when Q1+Q2=O
it will be :82500+63000+M2300000+50400=0
M2300000=-195900
M=-195900/2300000
M=-0.85
M=-0.1KG

With these
buttons you can write sub- and superscripts. Please use them...

I would agree with the 0.250 kg * ( 3.30 x 105 J/kg ) to melt the ice.
I don't understand what (3030*10^6) stands for

(4.2*103)*0.250*60 is the heat to bring the water from 0##^\circ## to 60##^\circ##, but: what do you do with the ice that has melted ?

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BvU said:
With these View attachment 240265 buttons you can write sub- and superscripts. Please use them...

I would agree with the 0.250 kg * ( 3.30 x 105 J/kg ) to melt the ice.
I don't understand what (3030*10^6) stands for

(4.2*10^3)*0.250*60 is the heat to bring the water from 0##^\circ## to 60##^\circ##, but: what do you do with the ice that has melted ?
The melted ice turns into water which will be heated and change into steam which is in gas form

That's not what the scenario describes: it says melted ice + water are heated to 60##^\circ##

BvU said:
That's not what the scenario describes: it says melted ice + water are heated to 60##^\circ##
the way i understoond it , its like the ice and water are in equlibrium, meaning they have same or equal temperature and they are now being heated at 60 degree celcius to a steam
or what is in your opinion?

It's not an opinion: it's what the problem statement describes.
You begin with
Julieh said:
0.250 kg of ice and 0.300 kg of water in thermal equilibrium
and they
are heated to 60°C

Julieh said:
which will be heated and change into steam
No, nothing is changed into steam. Steam is added to the ice and water mix, melting the ice and condensing the steam, with the whole ending up as water at 60°C.

Ha haru ! Bedtime for me , glad you're here !

haruspex said:
No, nothing is changed into steam. Steam is added to the ice and water mix, melting the ice and condensing the steam, with the whole ending up as water at 60°C.
now how is Q1 equation and Q2 equation look like?

BvU said:
Ha haru ! Bedtime for me , glad you're here !
i am still not getting it . but we can continue tommorow , have a blessed night

Julieh said:
now how is Q1 equation and Q2 equation look like?
The only problem with Q1 and Q2 individually is the erroneous 3030x106 in Q1 that @BvU mentioned.
The other problem is what you did with these two expressions. Q1 is the heat gained by the ice and water mix, while Q2 is the heat lost by the steam. What will be the relationship between those two quantities?

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