Calculating Heat Loss/Gain in Copper & Liquid

  • Thread starter Dark Angel
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In summary, the conversation is about a block of copper being heated and then dropped into a liquid. The questions posed include how much the copper cooled, how much energy it released, how much the liquid warmed up, and the specific heat capacity of the liquid. There is a hint given to consider the symbol for the liquid's specific heat capacity in order to determine the final temperature.
  • #1
hey guys i wuz thinking if anyone could help me here.

A block of copper of mass 10kg is heated to 100 degree C. it is then dropped into 2kg of a liquid at a temp.of 2o degree C.

a) thruough how many kelvin did the copper cool?
b) how much nrg did d copper release as it cooled?
c) by how many kelvin did d liquid warm up/
d) let the specific capacity of d liquid in j/kg/K be 'y'. in terms of 'y', how much nrg must the liquid have gained as it warmed up?
e) where did dis heat nrg gained by th liquid come from?
f) bearing in mind ur ans. to (b),(d) and (e), work out a value for d specific heat capacity of d liquid.
 
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  • #2
You know, there's a link to the "HomeWork" side at the topof this page ...

but I will hint that, if you only have a *symbol* for the liquid's "c",
the final temperature will *depend on* that symbol.

Tell me why ...
 
  • #3


a) The copper cooled from 100°C to 20°C, which is a decrease of 80°C or 80K.
b) The amount of energy released by the copper can be calculated using the specific heat capacity of copper (which is 385 J/kgK) and the change in temperature. So, Q = m x c x ΔT = 10kg x 385 J/kgK x 80K = 308,000 J.
c) The liquid warmed up from 20°C to 100°C, which is an increase of 80°C or 80K.
d) The amount of energy gained by the liquid can be calculated using the specific heat capacity of the liquid (which is y J/kgK) and the change in temperature. So, Q = m x c x ΔT = 2kg x y J/kgK x 80K = 160y J.
e) The heat energy gained by the liquid came from the heat energy released by the cooling copper.
f) By comparing the amount of energy released by the copper (308,000 J) and the amount of energy gained by the liquid (160y J), we can calculate the specific heat capacity of the liquid as y = 308,000 J / 160 J/kgK = 1925 J/kgK.
 

1. How do you calculate heat loss/gain in copper and liquid?

The formula for calculating heat loss/gain in copper and liquid is Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

2. What units should be used when calculating heat loss/gain?

The units used for calculating heat loss/gain in copper and liquid are joules (J) for energy, grams (g) for mass, and degrees Celsius (°C) for temperature.

3. How does the specific heat capacity affect heat loss/gain?

The specific heat capacity is a measure of how much heat energy is required to raise the temperature of a substance by 1 degree Celsius. A higher specific heat capacity means that more heat energy is required to change the temperature of the substance, resulting in lower heat loss/gain.

4. How does the mass of the substance affect heat loss/gain?

The mass of the substance directly affects the amount of heat energy required to change its temperature. A larger mass will require more heat energy, resulting in higher heat loss/gain.

5. Can the formula for calculating heat loss/gain be applied to other materials?

Yes, the formula can be applied to other materials as long as their specific heat capacity is known. Different materials may have different specific heat capacities, so the value of c in the formula will vary.

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