Problem Solving - Calculate the mass of steam condensed

AI Thread Summary
The discussion revolves around solving two physics problems related to heat transfer. The first problem involves calculating the mass of steam condensed when steam at 100°C is introduced to ice, resulting in a final temperature of 0°C for both the condensed steam and melted ice. The second problem focuses on determining the specific heat capacity of frozen vegetables placed in boiling water and the time required to reheat the mixture back to boiling after cooling. Participants emphasize the importance of showing work to facilitate understanding and suggest that the thermal capacity of the saucepan simplifies calculations. The conversation highlights common challenges in applying energy conservation principles in thermodynamics.
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Problem Solving -- Calculate the mass of steam condensed

1 ) A jet of steam at 100°C is directed into a hole in a large block of ice. After the steam has switched off, the condensed steam and melted ice are both a 0 degrees Celsius. The mass of water collected is 206 g. Calculate the mass of steam condensed.
Specific heat capacity of water is 4200 Jkg-1K-1
Specific heat capacity of ice is 2100 Jkg-1K-1
Latent heat of fusion is 330 X 1000 Jkg-1
Latent heat of vaporization is 2260 X 1000 Jkg-1

2 ) A mass of 450g of frozen vegetables is taken from a freezer at -20°C. The vegetables are immediately placed in saucepan containing 1100g of boiling water. The saucepan has a thermal capacity of 900J/K. The Final temperature of the saucepan,water and vegetables is 83°C.

a) calculate the specific heat capacity of the vegetables.

b ) the saucepan and its contents are then heated using a heater which provides 1200J of thermal energy each second. Determine how long it takes to bring the water back to its boiling point.

Can anyone solve these problems for me? Your help is very much appreciated.
 
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Alas, it seems that you have misunderstood the purpose of these forums. Was it not clearly explained when you signed up?
 
This looks like homework? If so it should be in the Homework section where you will get help to solve them yourself.
 
Can anyone solve these problems for me?
We can - but you will learn more if you do it, even if it is not homework.
Show me how you would approach these problems and I'll help you over the rough bits.

Also - it is easier to explain stuff to you if I can see how you think.
 
Energy Lost by water = Energy gained by vege + saucepan

4200X1.1X(1oo-87) = 0.45XcX(83+20) + m(900)

Im stucked at here. I use the equation mcθ.
 
seongwei said:
Specific heat capacity of water is 4200 Jkg-1K-1
Specific heat capacity of ice is 2100 Jkg-1K-1
Latent heat of fusion is 330 X 1000 Jkg-1
Latent heat of vaporization is 2260 X 1000 Jkg-1

2 ) A mass of 450g of frozen vegetables is taken from a freezer at -20°C. The vegetables are immediately placed in saucepan containing 1100g of boiling water. The saucepan has a thermal capacity of 900J/K. The Final temperature of the saucepan,water and vegetables is 83°C.

a) calculate the specific heat capacity of the vegetables.

b ) the saucepan and its contents are then heated using a heater which provides 1200J of thermal energy each second. Determine how long it takes to bring the water back to its boiling point.

I have no idea for question 1. For question 2 : Energy Lost by water = Energy gained by vege + saucepan

4200X1.1X(1oo-87) = 0.45XcX(83+20) + m(900)

Im stucked at here. I use the equation mcθ.
Welcome to physics forums. Now that you have included some working, we can see where you may be having problems.

I think it is more likely that the saucepan and its hot water are initially at the same hot temperature, so you should revise the equation in words. What would it be in this case?

The capacity of the saucepan is quoted as Joules/degree, so it seems that they have already taken into account its mass. So your calculations won't specifically need the saucepan's mass.
 
seongwei said:
Energy Lost by water = Energy gained by vege + saucepan
That would be for Q2.

Doesn't the pan and the water start out at the same temperature?
Therefore, shouldn't energy flow from the pan to the veges as well?

4200X1.1X(1oo-87) = 0.45XcX(83+20) + m(900)
Note: you are given the thermal capacity "C"(=900J/K) for the pan.
Thermal capacity is related to the specific heat "c" by C=mc.
 
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