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papz_007
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A 5.0kg block of copper is heated for 8 mins, by an electric heater of heat capacity 300jk-1 embedded in it. The potential difference across the heater is 25V and it carries a current of 2A. The temp. the block increases by is 10k.

a, calculate the specific heat capacity of copper.
b, Why is the calulated value greater than the correct value for copper.
c, How could you improve the experiment to obtain a better value.

Anyone answer this question for me please :aetsch:
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john !!
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the power supply has a power of P=IV = 25*2 = 50 W
this is 50 J per second. as its going for 8 minutes, that's 8*60 seconds; a total of 50*8*60 J delivered. that's 24000J.

first assumption: that all this energy is used to heat the copper and heater

the heat gain of the copper is equal to mass*shc*temp.rise = 50*c (call c the shc. we are looking for)

however, you have also given a shc of the heater (but not its mass!) so I'm not sure if you want me to model the heater as something being heated as well. if you do, I need it's mass, if you don't then you gave some irrelevant data. so I'll assume not.

24000 = 50c
c = 480 J Kg^-1 K^-1

(b) think about the assumption I made
(c) insulation
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nas7232
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A 5.0kg block of copper is heated for 8 mins, by an electric heater of heat capacity 300jk-1 embedded in it. The potential difference across the heater is 25V and it carries a current of 2A. The temp. the block increases by is 10k.

a, calculate the specific heat capacity of copper.
b, Why is the calulated value greater than the correct value for copper.
c, How could you improve the experiment to obtain a better value.

Anyone answer this question for me please

use

H=mc deta T
m = mass
c = specific heat capacity
T= change in temperature

rearrange equation -> c = H / mT

p=IV

bung in equation = answer for a)

b) the value is greater because enery was lossed due to lack of insulation, heat was used to warm up the block of copper intially, the thermometer does not touch the block where it was heated. All these are reasons of why the efficiency is not 100%. There are a few more too.

c) Insulate it better, put thermometer closer, use warmer copper so it don't use hit to warm up = wasted heat
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papz_007
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cheers pple

"by an electric heater of heat capacity 300jk-1 embedded in it"<<<< just wounderin y do they give u this reading for if i dont have to use it?
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elpaw
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you probably have to include the heat rise of the heater too.
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nas7232
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you do use it in part b)

your value will be greater then this value, the greater the difference between the two values = the less efficient the system is
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papz_007
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ah i see thanks man
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Ben C
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(Original post by mik1w)
however, you have also given a shc of the heater (but not its mass!) so I'm not sure if you want me to model the heater as something being heated as well. if you do, I need it's mass, if you don't then you gave some irrelevant data.
You're getting heat capacity and specific heat capacity confused. The OP gave a heat capacity, not a specific heat capacity.

Heat capacity already has the mass taken into account, hence the units J/K. Specific heat capacity is more general - it is the heat capacity PER KILOGRAM of the material. It therefore has units J/K/kg (compare with above). So you can use it for any size sample (assuming you know the mass).

See http://en.wikipedia.org/wiki/Heat_capacity for more.
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