# 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...

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!!!!!!!!!!!!!!!!!!!!!!

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### 4 Answers

Water

When 50g of water at 60oC are added to some water at 10oC, the temperature of the mixture is 15oC. What was the weight of the cold water?

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.

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.

1)

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 .

2)

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

3)

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

=696.5463 Joules

Let the resultant temperature be t.

Then the heat los

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

Similarly:

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

Answer:

Final temperature is 68.1818 degrees C.