Physics questions concerning Thermal energy, HELP PLZ!!!?

Question

I need help with this physics question:

A 4.0 kg bar of gold at 200°C is dropped into a tub that contains 50L of water at 20°C.
- What is the tempreature of the mixture when it come to equilibrium?
- How much heat did the gold bar lose?

any help is appreciated!

Answer 1

it is assumed that tub itself is not involved in the exchange of heat [delta -T (tub) =0] otherwise its data would be needed.

water in Tub:
volume of water = 50 litre
mass of water mw= 50 kg

let the equilibrium temperature by Tf, the conservation of heat energy gives

heast lost by gold = heat gained by water
mg* cg* [ Tg - Tf ] = mw* cw* [ Tf - Tw ]

where c - specific heat capacities

4*1291(J/kg-c)[ 200 -Tf ] = 50*4186(J/kg-c) [ Tf - 20]

5164 [ 200 -Tf ] = 209300 [ Tf - 20]

1032800 + 4186000 = 209300 Tf + 5164 Tf

Tf = 5218800 / 214464 =24.334 deg c

Heat lost by gold bar = 4*1291 [ 200 - 24.334]
= 907139.224 joule

Answer 2

You need to look up the specific heat of the gold (and water although that is just 1 cal/g deg C)

(Golds initial temp - final temp) x Aumass x Auspecific heat = energy transfered to water

(water's initial temp - final temp) x H2Omass x H2O specific heat = - energy transfered to water

If they come into thermal equilibreum, the two final temps are equal.

So you have 2 equations and 2 unknowns, solve for the final temp and the energy transfered to the water.

Good luck!

Answer 3

Let 't' be the equilibrium temperature.
We know,
heat lost by gold bar = heat gained by water
=> m[g] c[g] T[g] = m[w] c[w] T[w]

Where 'c' is the sybol for specific heat capacity of a substance & T is for change in temp.

m[w] = density x volume
=> m[w] = 1000 x 50/1000
=> m[w] = 50 kg

therefore,
4 x c[g] x (473 - t) = 50 x c[w] x (t - 293)

Once you have the values of the specific heat capacities of gold and water you can evaluate 't'.