HELP! How much heat energy would be transferred (multiple choice)?

Question

One contains water and the other contains an equal volume of another liquid that has half the density of water and two times the specific heat. You heat the water from 10 to 200C and the other liquid from 80 to 900C. What would describe the amount of heat energy transferred to raise the temperature of the other liquid and the amount transferred to raise the temperature of the water:

(a) Equal amount of heat energy is transferred to the water and to the liquid.
(b) The heat energy transferred to the liquid is half that transferred to the water.

Answer 1

@biire2u,

This thermodynamics problem was set up to trick you about the working fluid table.

The whole idea here is to teach a student about 1st law of thermodynamics, internal energy, work, enthalpy, solid state, compress liquid state,saturated liquid state, saturated state, saturated vapor state, Superheated states, and specific at each different state.

If you would apply 1st law of thermodynamics to this system.



Q(1-2) = U2 - U1 + PE + KE + W



Q(1-2) is heat transfer added to the system from state 1 to state 2

U2 is internal energy at state 2

U1 is internal energy at state 1

PE potential energy = Kinetic Energy = Work = 0

So,



Q(1-2) = U2 - U1 = (h + Pv) at state 2 - (h+Pv) at state 1

Where,

U = h + Pv

h = enthalpy

P = Pressure

v = specific volume



All the value h and v can be found on STEAM TABLE and pressure = atm (assumed P = atm pressure) All these values were done by scientist experiments for years.



For water, you can calculate it and get the answer Q(1-2).



BUT on the other hand, the working fluid in UNKNOWN and it's not an Ideal gas. How can you calculate enthalpy and specific volume at each state? It's a trick problem and can't be solved.



That explained why when solving thermodynamics problem, the 1st thing that you need to do is define your working fluid.

If it's a fluid ----> use table for the data

If a weird unknown fluid that does't have TABLE available on earth, don't bother to solve it because No one can solve it.

If it's Ideal gas--> use Ideal gas law Pv = nRT to obtain your data

Answer 2

c) neither?



The assumption is since the other fluid didn't mention anything about the phase change to a gas and you know that water has a gigantic absorption of heat in the phase change from water to gas.

1 kg of water absorbs 90C * 1000g = 90,000 calories of heat to go from 10C to 100C.



But for water to do a phase change it takes 540 calories PER GRAM to go from 100C liquid to 100C gas. So you have 1000g * 540 = 540,000 calories to do phase change.



And in this case you are raising the steam from 100C to 200C so more calories are needed to raise the steam temp. Specific heat of steam is about half of water so 1000g * 0.5 * 100C = 50,000 calories



So the heat added up to heat 1kg of water from 10C to 200C is:90,000 cal + 540,000 cal + 50,000 cal

= 680,000 calories to heat 1 kg of water.

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The unknown liquid has a specific heat twice the water, which will take twice the heat to raise the unknown liquid temp. 900C - 80C = 820 C temp differential



Unknown liquid has twice specific heat of water but since the density is half of water, then for the same volume you will have only 500g of the unknown liquid,so: 500g * 820C * 2 calories a gram = 820,000 calories



So the unknown liquid has 820,000 calories for 500g unknown liquid

and 680,000 calories for 1 kg of water .



This isn't either of your answers , but unless I did math error somewhere answer a)..... is closest

Answer 3

Please review your paraphrasing of the question.

Water increasing temperature from 10°C-200°C undergoes change

of phase. Does the other liquid?

Answer 4

seriously, do your own homework!