Calculating Heat Loss/Gain in Copper & Liquid

[Dark Angel]

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.

 

[lightgrav]

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 ...

 

[thepatientmental]

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.