Acombination of 0.250 kg of water at 20.0°c, 0.400 kg of aluminum at 26.0°c, and 0.100 kg of copper at 100°c is mixed in an insulated container and allowed to come to thermal equilibrium. ignore any energy transfer to or from the container and determine; a. the final temperature of the mixture. b. the change in the entropy of the universe in this experiment.

23.63 degree.

Explanation:

Specific heat of water = 4186 joule / kg degree

Specific heat of aluminium  = 900 joule / kg degree

Specific heat of copper = 386 joule / kg degree

Let equilibrium temperature be T.

Heat will be gained by water and aluminium and lost by hot copper.

Heat gained or lost = mst , where m is mass , s is specific heat and t is rise or fall of temperature.

Heat gained by water = .250 X 4186 X ( T - 20 )

= 1046.5 ( T-20)

Heat gained by aluminium = .400 x 900 x ( T - 26 )

= 360 ( T - 26 )

Heat lost by copper = .100 x 386 x ( 100 - T )

38.6 ( 100 - T)

Heat gained = Heat lost

1046.5 ( T-20) + 360 ( T - 26 ) = 38.6 ( 100 - T)

T ( 1046.5 + 360 + 38.6 ) = 3860 + 9360 +20930

1445.1 T = 34150.

T = 23.63 degree.

Change in the entropy of the universe will be zero because no heat is exchanged with the universe. The container is insulated from outside .

Inside the container,  entropy will be increased.  

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .