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anonymous
 3 years ago
The specific heat capacity of water= 4200 J kg1K1
1) How much heat energy is required to warm a) 1 kg of water from 14°C to 15°C; b) 5 kg of water from 14°C to 15°C; and c) 5 kg of water from 14°C to 22°C. THANK YOU!!
anonymous
 3 years ago
The specific heat capacity of water= 4200 J kg1K1 1) How much heat energy is required to warm a) 1 kg of water from 14°C to 15°C; b) 5 kg of water from 14°C to 15°C; and c) 5 kg of water from 14°C to 22°C. THANK YOU!!

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[J kg ^{1} K ^{1}\] this is the correct way of writing the unit

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2Use the formul heat required =\(mc \Delta t\)

AravindG
 3 years ago
Best ResponseYou've already chosen the best response.2where m is mass c is specific heat capacity ,delta t=change in temperature

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i worked it out...is the answer 4200J

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm and b) is 21000J while c) is 1.68*10\[1.68\times10^{5}\] am i right???

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0question 2! one way to measure the energy contained in a beam of xrays is to absorb the beam by a piece of pure graphite. If xrays are shone on to a cavity in a block of graphite of mass 10 g (0.01 kg), and a rise in temperature of 0.005K, work out the energy supplied by the beam.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh and the specific heat capacity of graphite is 600\[J kg ^{1}K ^{1}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok......just making sure. question 3! An electric drill consuming energy at a rate of 200 W is used to make a small hole in a block of copper of mass 2 kg. Assuming that 80% of the electrical energy used ends up in the block as heat, and that this heat is spread evenly throughout the block, estimate its rise in temperature in 10 seconds.

austinl
 3 years ago
Best ResponseYou've already chosen the best response.0Why dont you make a new question? rather than lump several into one post?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok @kym02 i'll give a brief info. so energy is 200 W i.e. it uses 200 J in 1 second, correct?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0um i honestly dont know....im a beginner. Dont really know about this stuff! sorry!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok, then there is one concept. W means Watt. Watt is the unit of power of a machine. We say that the power of a machine is the amount of work it can do in 1 second.(i.e. rate of doing work,) Hence its unit is Watt(W) also expressed as Joule/second. so 200 W basically means that the machine will use 200 Joules of energy in 1 second. Ok?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Fine. so if the machine uses 200 Joules of energy in 1 second, how much of energy would it use in 10 seconds?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok. 2000 J is the electrical energy used. Now he says that 80% of this 2000 J is used as heat energy. So what is 80% of 2000 ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thats right. So this 1600 is the heat energy. As shown above, Heat energy = Mass x sp. heat capacity x rise in temp. You have mass = 2 kg, sp. heat capacity(c) = 400 JKg^1K^1, can you find the rise in temperature?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i dont know that one, do i have to transpose?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yep you have to. above there was a formula mentioned Heat energy = m x c x delta T Here Heat energy = 1600, m = 2 kg, c = 400 J/KgK, u need to find delta T

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0m means Mass; c means specific heat capacity and delta T means the rise in temperature.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[E _{H}=mcΔT\] so the temperature rise will be: EH divided by mass times c am i right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thats perfectly correct. Just type d ans also when you find it. :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok i will, im working on it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes!!!!! thanks :) i have one more question though

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a)How much potential energy does a 2.0 kg lump of copper possess when it is lifted 3.0 m from the ground? (Take g to be 10 N kg^1.) b) If it is dropped, how much kinetic energy will it have just before it hits the ground? c) If all that energy is converted to heat, and if all that heat goes into the copper, what rise in temperature would you expect? That's all ! :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a) m = 2, h = 3, g = 10. Use Potential Energy = mgh. b) Potential Energy found in a) = Kinetic Energy (By law of conservation of energy) c) Whatever PE you get, assume that as heat energy, the use C , M of copper to find rise in temp by formula EH = m.c. delta T.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0whats the unit for a) ???

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0a) Any form of energy, whether potential energy, electric energy, heat energy, is expressed in joules( J ) (Please please note this fact; )

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and c) = \[7.5\times 10^{2} \deg C\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I hope you're familiar with law of conservation of energy right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no but i will just look back in a textbook! thank you soooo much!!! you saved me!! :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah no probs as long as you're familiar with concepts and numericals :)
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