Help with internal energy Physics A level OCR
A heater has a constant output power of 6.0kw. The water is heated as it travels through the heater. The internal cross-sectional area of the pipe is 4.9*10^3 J kg^-1 K^-1/
(i). Calculate the temperature of the water at which it leaves the heater.
So i know that e=mc*temp but how would I work out the mass here?
The markscheme multiplies the cross sectional area by 0.200 and 1000 (which is the density of water so thats fine) - but I don't understand where this 0.2 value came from... If anyone can help me out please thank you.
(i). Calculate the temperature of the water at which it leaves the heater.
So i know that e=mc*temp but how would I work out the mass here?
The markscheme multiplies the cross sectional area by 0.200 and 1000 (which is the density of water so thats fine) - but I don't understand where this 0.2 value came from... If anyone can help me out please thank you.
Original post by MrToodles4
A heater has a constant output power of 6.0kw. The water is heated as it travels through the heater. The internal cross-sectional area of the pipe is 4.9*10^3 J kg^-1 K^-1/
(i). Calculate the temperature of the water at which it leaves the heater.
So i know that e=mc*temp but how would I work out the mass here?
The markscheme multiplies the cross sectional area by 0.200 and 1000 (which is the density of water so thats fine) - but I don't understand where this 0.2 value came from... If anyone can help me out please thank you.
(i). Calculate the temperature of the water at which it leaves the heater.
So i know that e=mc*temp but how would I work out the mass here?
The markscheme multiplies the cross sectional area by 0.200 and 1000 (which is the density of water so thats fine) - but I don't understand where this 0.2 value came from... If anyone can help me out please thank you.
question seems garbled and lacking in information
4.9*10^3 J kg^-1 K^-1/is not in units that can be an area, areas always have units that are measures of length squared.
please try again or post a link to (or photo of) the original question.
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