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What is the final temp when 10grams of water at 0*C is added to 100 grams of water at...
What is the final temp when 10grams of water at 0*C is added to 100 grams of water at 75*C?
i also had 2 more questions i wasn't sure of...
What is the final temp when 10 g of ice at 0*C is added to 100g of water at 75*C? (Hf=6.0kJ/mol, heat capacity of water is 4.2 J/g*C)
Calculate the amount of energy in Joules required to change 10g of solid mercury at its melting point to mercury vapor at the boiling point. The mp, bp, and specific heat of mercury are -39*C, 375*C, and 0.140J/g*C, respectively. Compare with the amount of heat needed to change 10g of ice at 0*C to steam at 100*C (heat of fusion Hg=11.4J/g, heat of vaporization Hg=5.91 x 10^4 J/mol)
the explanation to any one of the questions will be GREATLY APPRECIATED!!!!!!!!!!!!!!!!!!!!!!
3 Answers | add yours
Let us say that the final temperature of mixture of 10 grams of water at 0 degrees C, and 100 grams of water at 75 degrees C is t. Then:
Heat gained by water at 0 Degrees C
= (Weight of water)*(Final temperature - Initial temperature)
= 10*(t - 0) = 10*t
Heat lost by water at 75 Degrees C
= (Weight of water)*(Initial temperature - Final temperature)
= 100*(75 - t) = 7500 - 100 t
But heat lost is equal to heat gained. Therefor:
10t = 7500 - 100t
Therefor: 110t = 7500
Therefor: t = 7500/110 = 68.1818 degrees C
Final temperature is 68.1818 degrees C.
Posted by krishna-agrawala on January 21, 2010 at 12:26 PM (Answer #1)
High School Teacher
We use the principle that heat geained by the lower temperature object = heat lost by the higher temperature object at the resultant temperature and no heat is lost to the surrounding.
When 10 grams of water at 0 C is added to 100 grams of water,
10 gram of water gains the heat of 10(t-0)s which is equal to the heat lost by the 100 gms of water= 100s(75-t).Here t is the resultant temperature and s is the specific heat of the water. So 10s t = 100s(75 - t) or 110t= 100*75 or t =100*75/110 = 68.1818 C .
Gain of temperature by 10 gm ice = 10*6000/18*(4.2))+10(t-0) calories = loss of heat 100(75-t) calories by 100 gram water.Solving for t, we get: 110t = 7500 - 60000/(18*4.2) .So t = [7500 - 60000/(18*4.2)]/110 = 60.9668 C
The heat required to raise the temperatutere of mercury from melting point to vapouristion = Heat required to melt it to its at mp + heat required to raise the temperature from mp to bp + heat required to vaporise.
= 10*11.4 +10*[375-(-39)]*(0.14)+10(5.91/200.59)
=114+579.6 + 2.9463
Let the resultant temperature be t.
Then the heat los
Posted by neela on January 28, 2010 at 3:25 PM (Answer #2)
High School Teacher
hot = cold
Mh(ti - tf) = Mc(tf - ti) where Mh is the mass of the hot water, Mc is the mass of the cold water, Tf is final temp and Ti is the initial temp of the water.
For mercury you are going through phase changes so...
Hf*Mass = joules to melt
spht * mass * delta T = Joules to heat
Hv * mass = joules to vaporize
Add these values to gether to get your answer.
Posted by ellyril1 on April 10, 2010 at 1:24 AM (Answer #3)
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