Help calculating the equilibrium temperature?
Elemental sulfur exists in two crystalline forms, rhombic and monoclinic. From the following data, calculate the equilibrium temperature at which monoclinic sulfur and rhombic sulfur are in equilibrium. ∆Hf° (kJ/mol) S° (J/K . mol) S (rhombic) 0 31.88 S (monoclinic) 0.30 32.55
Help!!! calculate final temperature!!!?
You would think more people would answer these but I saw it at the bottom of the page so I grabbed it. Q1 = heat lost by liquid 1 Q2 = heat gained by liquid 2 Q = mc(delta T) Q1 = Q2 m1c1(Thot - Teq) = m2c2(Teq - Tcold) specific heat (c) is the same for both since they're both water so you can remove it from the equations. m1(Thot-Teq) = m2(Teq - Tcold) 40g(65C - Teq) = 50g(Teq - 10C) 2,600g-C - 40gTeq = 50gTeq - 500g-C 3,100g-C = 90gTeq 3,100g-C/90g = Teq 34.44C = Teq .
Chemistry help!!!! Calculate temperature rise...URGENT!?
The potential energy of water at the top of a dam or waterfall is converted into heat when the water dashes against rocks at the bottom. The potential energy of the water at the top is equal to Ep=mgh, where m is the mass of the water, g is the acceleration of the falling water due to gravity (g = 9.81 m/s2), and h is the height of the water. Assuming that all the energy is converted to heat, calculate the temperature rise of the water in degrees Celsius after falling over California’s Yosemite Falls, a distance of 739 m. The specific heat of water is 4.18 J/(g⋅K). delatT = _______ºC I don't know what to do with this problem because they don't give a mass amount, hep please!!
Calculating temperature change?! please help?
Calculate the temperature change for mercury if 160 grams of the metal absorbs 1500J of heat energy. Mercury's specific heat is .14J/(g x degrees C) please help me with this chemistry problem!
Calculate the freezing temperature depression. Help!?
Wow, your teacher is really trying to throw you off here! This is a tricky one -- I had to read it a few times. So as you said >>> deltaT=(constant)(molality) and >>> (constant) = 40*CKg/mol So to find deltaT, all you need now is the molality of the solution! Everything else (like the melting point) is just there to confuse you. (molality)=(mol solvent)/(Kg solution) Camphor = C10H16O Hopefully you can get it from here. If not, feel free to email me (through my YA profile) and I'll be glad to help you some more. :)
Help with calculations?
If the average or normal temperature decreases with altitude in the troposphere is 6.5degreesC/km, calculate the approximate altitude in which a pilot would expect to find each of the following atmospheric temperatures, if the surface temperature is 27degreesC. 10degreesC: _____meters 0degreesC:______meters Thanks.. i dont know how to work this out!!
How do I calculate weight of water at different temperatures?
The weight of a given mass of water does not change with temperature. But, yes, if you are talking about volume ... and you seem to be, the density of water changes about 5% in its liquid form over temperature. And at 4C, it is its most dense. I need to calculate this myself, from time to time, so I generally type something like "Water density vs. temperature" into a search engine (I use Bing) and I can usually find a chart. Look: here is the first result from Bing Water - Density and Specific Weight This helps me calculate the weight of a volume of water at different temperatures.If there is not enough resolution in the chart, search for a better one, or interpolate.
How can we calculate critical temperature, volume and pressure in terms of a and b?
This can be done from the Van der Waals Equation of State for real gases. Consider that equation, for one mole of the gas in question:[math](P+\frac{a}{V^2})(V-b)=RT[/math]A bit of rearrangement gives us-[math]P=\frac{RT}{V-b}-\frac{a}{V^2}[/math]Now if a graph of P as a function of volume were to be plotted, it would be something like this-As you can see, the critical temperature [math](T_c)[/math] also happens to be a point of inflection for the PV diagram. One concludes, at this temperature-[math]\frac{\partial P}{\partial V} = 0[/math][math]\frac{\partial^2 P}{\partial V^2}=0[/math]Where P is a function of V as per the Van der Waals' Equation. Making use of that, we get:[math]\frac{\partial P}{\partial V} = -\frac{RT}{(V-b)^2} + \frac{2a}{V^3}=0[/math][math]\Rightarrow \frac{2a}{V^3} = \frac{RT}{(V-b)^2} . . . . . . (A_1)[/math][math]\Rightarrow \frac{a}{V^4} = \frac{RT}{2V(V-b)^2} . . . . . . (A_2)[/math]Also, by differentiating the first partial derivative again, with respect to V, we get:[math]\frac{\partial^2 P}{\partial V^2} = \frac{2RT}{(V-b)^3} - \frac{6a}{V^4} = 0[/math]Dividing both sides by 2, and taking one term to the left hand side, we get:[math]\frac{RT}{(V-b)^3} = \frac{3a}{V^4}[/math]Plugging in the value that we just obtained from Equation [math]A_2[/math]-[math]\frac{RT}{(V-b)^3} = \frac{3RT}{2V(V-b)^2}[/math]Now, if we divide both sides by a factor of [math]\frac{RT}{(V-b)^2}[/math], and just rearrange the terms, we see that-[math]3V - 3b = 2V[/math][math]\Rightarrow V_c = 3b, [/math]which is the critical volume.Plugging this value into equation [math]A_1[/math]-[math]\frac{RT}{4b^2} = \frac{2a}{27b^3}[/math][math]\Rightarrow T_c = \frac{8a}{27Rb},[/math]which is the critical temperature.To calculate the critical pressure, return to the original equation-[math]P=\frac{RT}{V-b}-\frac{a}{V^2}[/math]Plug in the values of [math]T_c[/math] and [math]V_c[/math] we just calculated, and you'll see that-[math]P_c =\frac{a}{27b^2}[/math]Image Source: Bright Hub Engineering.
Calorimetry calculations help?
First, you calculate how much heat went into raising the temperature of the water. This is done with the formula: Q=mc delta T hence Q=(100)(4.18)(64.9) Q=27128.2 J Now we find how many moles of propane was burned. First we calculate the molar mass of propane 3(12.01)+8(1.01)=44.11 g/mol You burned .6 grams, so this is .6/44.11=.0136 mol The standard enthalpy of combustion is defined by the heat released per mole. So all we do is divide how much energy was released by the number of moles we have. 27128.2/.0136=1994721 J/mol usually, this is expressed in kJ however, so 1995 kJ/mol hope this helped