Can anyone show me how to do this past examphysics question that I'm struggling with?
He plunges a red hot iron poker, initially at 9000C, into an insulated bowl containing 1 litre of the fermented grape juice, initially at 100C. What temperature will the wine end up at? (You should assume the following: specific heat capacities of wine [mostly water] and iron are 4200 Jkg-1K-1 and 450 Jkg-1K-1 respectively; mass of poker, 0.3 kg.)
Thanks! xx
He plunges a red hot iron poker, initially at 9000C, into an insulated bowl containing 1 litre of the fermented grape juice, initially at 100C. What temperature will the wine end up at? (You should assume the following: specific heat capacities of wine [mostly water] and iron are 4200 Jkg-1K-1 and 450 Jkg-1K-1 respectively; mass of poker, 0.3 kg.)
Thanks! xx
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Q=mcT and the sum of the Q's must equal 0
Q is energy transfer, m is mass, c is specific heat, and T is change in temperature.
the change in temperature is the equilibrium temperature - the initial temperature.
the iron poker has a Q and the wine has a Q (the poker gives it's Q to the wine)
we assume the wine is water so 1 liter of water weighs 1 kilogram.
so mcT+mcT=0
(.3kg)(450 J/(kg*C))(T-900C)+(1kg)(4200 J/(kg*C))(T-10C)=0
we're solving for the equilibrium temperature the poker and wine will meet at. without the units this equation is:
.3*450*(T-900)+4200(T-10)=0
use the distributive property to simplify.
(130T-121500)+(4200T-42000)=0
combine like terms
4330T-163500=0
solve for T
4330T=163500
T=37.76 degrees Celsius
The iron poker will cool down and the wine will heat up until the meet at 37.76 degrees Celsius.
you can see how the iron is much better conductor than water. It cools down about 850C and the wine only increases by about 27C.
Q is energy transfer, m is mass, c is specific heat, and T is change in temperature.
the change in temperature is the equilibrium temperature - the initial temperature.
the iron poker has a Q and the wine has a Q (the poker gives it's Q to the wine)
we assume the wine is water so 1 liter of water weighs 1 kilogram.
so mcT+mcT=0
(.3kg)(450 J/(kg*C))(T-900C)+(1kg)(4200 J/(kg*C))(T-10C)=0
we're solving for the equilibrium temperature the poker and wine will meet at. without the units this equation is:
.3*450*(T-900)+4200(T-10)=0
use the distributive property to simplify.
(130T-121500)+(4200T-42000)=0
combine like terms
4330T-163500=0
solve for T
4330T=163500
T=37.76 degrees Celsius
The iron poker will cool down and the wine will heat up until the meet at 37.76 degrees Celsius.
you can see how the iron is much better conductor than water. It cools down about 850C and the wine only increases by about 27C.
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Since the bowl is insulated, we can ignore the heat absorbed by the bowl as negligibly small.
Latent heat of steam = 2260 000 J/kg
Melting point of iron is nearly 1500° C
Hence the temperature given in the question is wrong and we can take it as 900 °C
The quantity of heat given out by the iron poke if its temperature is brought down to 100° C is 0.3*450*(900 – 100) = 108000 J
2260 000 J of heat is required to convert 1 kg of water at 100°C into steam at100°C
108000 J of heat will convert 108000 / 2260 000 = 0.047 kg of water.at 100°C into steam at100°C
Hence the temperature of iron poke will be reduced to100°C and 0.047 kg of water at 100°C will be converted into steam at100°C
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Latent heat of steam = 2260 000 J/kg
Melting point of iron is nearly 1500° C
Hence the temperature given in the question is wrong and we can take it as 900 °C
The quantity of heat given out by the iron poke if its temperature is brought down to 100° C is 0.3*450*(900 – 100) = 108000 J
2260 000 J of heat is required to convert 1 kg of water at 100°C into steam at100°C
108000 J of heat will convert 108000 / 2260 000 = 0.047 kg of water.at 100°C into steam at100°C
Hence the temperature of iron poke will be reduced to100°C and 0.047 kg of water at 100°C will be converted into steam at100°C
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