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Assuming no heat losst by the system what will be the final temp. when 11.2 g of water at 10.0 deg. C are mixed with 50.8 g of water at 55.0 deg C?
Wiki User
∙ 13y ago
you can't.
You can calculate the heat released, and then convert that.
Hmmm...how about:
11.2*10=112 gC
50.8*55=2794 gC
112+2794 = 2906gC
2906/(11.2+50.8)= 46.87C
Quick check: 10C < 46.87C < 55C the answer is also closer to 55C because of the greater mass (almost 5x) of the hotter water.
Anyone see that as correct? :)
Wiki User
∙ 13y ago
The final temperature can be calculated using the principle of conservation of energy, which states that the total heat lost by the hot water equals the total heat gained by the cold water. By applying this principle, we find that the final temperature will be approximately 47.2 degrees Celsius.
Wiki User
∙ 13y agoTwo q's to 0 problem.
q(Joules) = mass * specific heat * change in temperature
(50.0 g H2O)(4.180 J/gC)(Tf - 10.0 C) + (10.0 g H2O)(4.180 J/gC)(Tf - 50.0 C) = 0
209Tf - 2090 + 41.8Tf - 2090 = 0
250.8Tf = 4180
Temperature Final = 16.7 Celsius
--------------------------------------------
Wiki User
∙ 13y agoThe second temperature initial is not there. Here is set up:
(mass H2O)(specific heat H2O)( Tf - Ti) + (mass H2O)(specific heat H2O)(Tf - Ti) = 0
Where mass of H2O can be taken as the same as its volume in grams;
Tf is the final temperature in ºC (because there is a change in temperature, it does not have to be converted to Kelvin)
Ti is the initial temperature in ºC
specific heat of water = 4.18 J /K/g
Wiki User
∙ 13y agoSorry the question is unanswerable as we do not know the temperature of the ice.
Wiki User
∙ 11y ago109kj
Wiki User
∙ 13y ago30 deg C.
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