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?

Answer 1

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? :)

Answer 2

Two 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

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Answer 3

The 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

Answer 4

Sorry the question is unanswerable as we do not know the temperature of the ice.

Answer 5

109kj

Answer 6

30 deg C.