A100 gram cylinder of copper initially at temperature of 200 degrees celsius immersed in 400 grams of water which is at 20 degrees celsius. find the equilibrium temperature of copper and water system? specific heat of copper is 387 j/kg degrees of copper

temperature of water is 24 degree

temperature of copper is 26.89 degree

Explanation:

given data

copper mass = 100 gram

temperature = 200 degree

water mass = 400 gram

temperature = 20 degree

specific heat of copper = 387 J/kg = 0.387 J/g

to find out

temperature of copper and water

solution

we know in equilibrium condition heat is loss by copper is equal to heat is gain by water and we consider here x is temperature so equation is

mcΔt for copper = mcΔt for water    1

100 (0.387) (200-x) = 400 (4.187) ( x-20)    2

solve it and we get x

x = 24.06

so here temperature of water is 24 degree

and put this value in equation 2 for water we get temperature of copper

100 (0.387) (200-x) = 400 (4.187) ( x-20)

100 (0.387) (200-x) = 400 (4.187) ( 24-20)

solve it and we get

x = 26.89 degree

so temperature of copper is 26.89 degree

inverse v=i*r

nice nice