Specific Heat and Calorimetry questions

Question

Need some help with these questions for online schooling, and don't know what formula to use or how to do the questions in general. Any assistance is appreciated, thanks in advance.

1.      A calorimeter has a heat capacity of 4.18 kJ/g ^oC. Complete combustion of 1.00 g of hydrogen in this calorimeter causes the water temperature in the calorimeter to increase 3.54 ^oC. If the liquid in the calorimeter has a mass of 9.569 g, calculate the molar enthalpy of combustion for hydrogen from this evidence.

2.      A reference gives the molar enthalpy of combustion for methane as -803 kJ/mol. What minimum mass of methane must be burned to warm 4.00 L of water from 22.4 ^oC to 87.6 ^oC, assuming no heat losses?

3.      Combustion of 3.50 g of ethanol, C₂H₅OH_(l), in a calorimeter with a heat capacity of 1.38 kJ/g ^oC causes a temperature increase of an unspecified liquid to rise from 19.88 ^oC to 26.18 ^oC. If the volume of the liquid is 11.0 mL, find the molar enthalpy of combustion for ethanol from this evidence.

4.      Find the temperature increase expected for 1.00 L of water when it absorbs all of the energy from the combustion of 1.00 g of acetylene, C₂H_2(g). The molar enthalpy of combustion for acetylene is -1.29 MJ/mol.

Answer 1

1.) Here you want to use q=m*C*∇T (That triangle is upside down but lets pretend it's not)

Where m is mass, C is specific heat, and T is change in temperature.

Also I believe the specific heat is J/g*C not kJ

We want to find how much heat the combustion of hydrogen gave off per mole of hydrogen.

So if this combustion of hydrogen happens inside of a calorimeter, the calorimeter allows us to measure how much heat was given off by use of the equation above. So we can plug in the mass of water that absorbed the heat in the Calorimeter, as well as the specific heat of water and the change in temp of the calorimeter

q=(9.569g)*(4.18J/g C)*(3.54) This problem gives us change in heat but sometimes you have to do final heat-intial heat

q=141.59 J This is the heat that the burning of hydrogen gave off.

This is the amount of heat given off for one gram of hydrogen burning, lets change it to how much heat is evolved per mole of hydrogen. We do this by using the molar mass to covert grams to moles in the denominator.

(141.59 J/gram H₂) * (2.016 grams H₂/mol) = 285 J/mole Final answer

2.) We wanna use q=m*C*∇T again, Basically find the amount of heat needed to make the water undergo this temperature change, and then find how much methane one needs to do this, to get delta T we have to do final-initial, we can use the density of water to covert from Liters to Grams, the density of water is just 1 gram/mL so....

(4000 mL) * (1 g/ mL) = 4000 grams Water

q = (4000 grams) * (4.18 J/ g C) * (87.6-22.4)

q=1.090 *10^6 J (that's a lot of heat, but we have lots of water)

now we can use the molar heat of combustion to find how much methane we need. First convert to kJ...

(1.090*10^6 J) / (1000J/kJ) = 1090 kJ

(1090 kJ) / (-803 kJ/mole) = 1.357 moles

convert to grams, using molar mass of methane

(1.357 moles) * (16.04 grams/mole) = 21.77 grams methane.

(Check your question, I think the specific heat of water has the wrong units, I made some corrections to the first problem)

3.) For this one I believe we would need more information, like the density of the liquid in the calorimeter, once we have this it is basically identical to the first problem.

4.) For this one we need to find the heat that this much acetylene will produce and then see the temperature change it will cause, so basically we find everything but ∇T in the q=m*C*∇T solve for it, and that is our answer.

First convert to moles of acetylene

(1.00 gram acet) / (26.04 g/mol) = .03840 moles acet

then find the heat this will evolve

(.03840 moles) * (-1.29 kJ/mole) = .04953 kJ minus sign doesn't matter here basically minus just tells us it is an exothermic reaction and will give off heat, but we just want to plug in the absolute value here

49.53 J = (1000grams)* (4.18g/J C) * ∇T Be sure to keep units consistent

∇T = 0.0118

Answer 2

Hi Wyatt Edward J.,

I'll discuss the first problem, and perhaps you can use that to figure out the rest.

The calorimeter heat capacity means, if you put in 4.18 kJ of heat, it will raise the temperature of 1 g of the thermal "ballast" liquid contents by 1 degree C. Then if you put in twice that amount of heat, you could either raise 2 g of the liquid by 1 degree C, or raise 1 g of liquid by 2 degrees C, or any combination of liquid mass and temperature which also multiplies to 2 . Pause for a moment and think about that, and play with some multiples of your own choice, until you thoroughly understand what the heat capacity value means. Another way to think about this, is, imagine you are heating some water on a stove. The flame is supplying heat at a constant value. If you have only a little water in your pot, it will heat up fast; but if you have a full pot, it will take much longer. Either way, you have put the same amount of heat into the water! The water in your pot, is just the same as the liquid in the calorimeter, in terms of the cause/effect relationship of heat/amount/temperature_change.

Now let's get back to your actual problem. You heated 9.569 g liquid -- so that's 9.569 times harder to heat than is 1 g liquid -- by 3.54C, that's 3.54 times harder than heating it just 1C. So put those together by mutiplying (since thay are multiplied in the calorimeter heat constant units of measure), that's 33.874+ times harder than heating 1 g by 1 C. Ok so far? Now, how much heat would the 1g, 1C condition have taken? that's the calorimeter heat constant, 4.18 kJ : put that all together, and you find that you must have produced 141.59 kJ of heat to do what you did.

Now, that was from 1.00g hydrogen -- but, that is only (1/2) of a mole of hydrogen gas (remember, molar heat of combustion refers to a mole of stuff in its standard state, which is the diatomic gas, for hydrogen). So, double your figure, to calculate for 1 mole of hydrogen gas.

So to recap, you need the formula: heat produced = (calorimeter constant) * (g calorimeter fluid) * (temperature change) .

I might point out, that real calorimeters sometime have a heat constant expressed as (a + b * m), where a refers to the contribution of the vessel walls, etc. and b refers to the heat capacity of the filling liquid of mass m). When combustions, etc. are done under oxygen-pressurized conditions, it requires a robust, metallic combustion chamber, which certainly has its own calorimetric heating requirement. Also, in those conditions, there's also a bit of heat contributed by the initial device which sets off the reaction, typically a small resistive wire loop which is heated electrically to flash burn it. That gets subtracted right off, before other calculations are done. Just sayin'.

-- Cheers, -- Mr. d.