Class+Notes

===__The Calorimetry Mini Problem Set__! Here, you will nominate a scribe from each group who will write a description of how to find the solutions to problems 1-7. The answer itself is already on the little sheet.===

__**Givens:**__
 * Question Number || Descriptive Answer ||
 * 1 || The first thing to do divide your givens in to two categories: the givens for Cu and the givens for H2O.

m= 1000g = 1 kg T1= 20 degrees Celcius T2 =80 degrees Celcius c= 0.385 kJ/kg (degrees Celcius)
 * Cu**

m= 500g = 0.5 kg T1= 20 degrees Celcius T2 =80 degrees Celcius c= 4.18 kJ/kg (degrees Celcius)
 * H2O**

you use this information to find Q(Cu) and Q(H2O) using our favourite equation Q=mc(delta T) and then you add the two values of Q for your final answer


 * remember**: the initial and final temperatures are the same for both substances because they are sharing a common environment ||
 * 2 ||  ||
 * 3 || __Given__:
 * CH4's MM= 16.04 g/mol
 * Vw=100L---> Mw=100kg=10^5g
 * 1 mole of methane=900KJ=9*10^5J

Qw = (10^5)(4.18 J/g*C)(45-20*C) Qw = 10450000J=1.045x10^7J --->Qmethane= -1.045x10^7J
 * Qw = Mw x Cw x delta T**

1 mole CH4 **:** 9x10^5= X mole CH4 **:** 1.045x10^7 X mole CH4=11.61 mol

M = MM x N M = (16.04 g/mol) (11.61 mol) M = 186.24 g of methane


 * therefore, there are 186.24g of methane must be burned to heat 100L of water. ||
 * 4 ||  ||
 * 5 || **__Qw = Mw x Cw x delta T__**

Msub = 1g Vh2o = 50ml delta T = ? Mw = 50g Qw = 25.8kj/mol

MM (NH4NO3) = 80g/mol --> Moler Mass N (NH4NO3) = 1g divided by 80g/mol > Moles = 0.0128mol

0.0128 mol x 25.8 Kj = .3225 Kj .3225 Kj = 322.5 J -> Qw value

322.5J = 50g x 4.18J/g x Delta T 322.5J divided by 50 divided by 4.18 = Delta T Delta T (approx) = 1.5 degrees Celsius || Q=MCT, the heat capasity of water is 4.2 J/g.c and the intial temprature is 30.0 c, and the final temprature is 31.5 c. So the change in temprature is 1.5 c. part A asks for the **Rate of heat given off** Using the formula, Q = (60,000 g (remember that it should be in grams) * 4.2 J/g.c * 1.5 c = 376200 J = 376.2 KJ. since all the given values we have is respect to the __enthalpy in one day__, we would say Q/ 1 hour = 376.2 Kj/ hour. The rate at which the person is giving off heat in one hour is 376.2 kj/hour. Part b is asking for the amount of heat given off in a day (a day is 24 hours) so there is a simple multiplication. 376.2 kj/hour * 24 hour (notice that the "hour" will cancel each other out) = 9028.8 kj of energy. the answer with respect to Sig Fig is going to be 9.03 * 10^3 kj ||
 * 6 ||  ||
 * 7 || The first thing you need to remember is that 1.0 L = 1.0 kg, since we can not use volume in the formula



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