A) In an insulated system, 61.0 grams of ahot metal at 600 degrees C was added t

Question

A) In an insulated system, 61.0 grams of ahot metal at 600 degrees C was added to 800 g water at 20.00 degrees C. The final temperature of the mixture became 29.38 degrees C. Calculate the specific heat of the hot metal. c(H2O)= 4.184 J/g

B) if 25 grams of liquid water at 0 degrees C is mixed with 75 grams of liquid water at 100 degrees C, what is the final temperature of the resulting mixture? Assume the mixing is carried out in an insulating container and no heat is absorbed by the container.

Thanks!

Explanation / Answer

A) heat lost by metal = heat gained by water

MCdt= mc'dt'

Where

M = mass of hot metal= 61.0 g

C= specific heat capacity of metal= ?

dT= Change in temperature of metal= 600-29.38= 570.62 oC

m= mass of water= 800 g

c= specific heat capacity of water= 4.184 J/ goC

dT'= chaCha in temperature of water= 29.38-20.00= 9.38 oc

Plug the values we get C= 0.902 J/goC

B) let t be the common temperature attained by the system and is in liquid phase.

Then heat gained by water at 0 oC= heat lost by water at 100oC

MCdt= mCdt'

Mdt= mdt'

25 g*(t-0) = 75 g*(100-t)

t= 75.00 oC

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