25

They're not exact numbers. These numbers aren't exact for three reasons: Each type of carb, protein, and fat has a different caloric value. These are overall averages for each class. Even if you were dealing with a single pure compound, the value couldn't be exact because there is individual variation in how much of that compound is metabolized based on ...


10

It doesn't "take one calorie to burn", you gain one calorie by burning. Now noting obviously what everybody else has said about selection of the type of fat affecting the precise amount of energy available from one gram (which in practice will be measured in a calorimeter), but what this is really saying is that by metabolising some specified ...


5

All you did is essentially right, your only mistake is in the last step, as LDC3 already pointed out in the comments. However, I am encouraging you to use units all the way and when dealing with thermodynamics use Kelvin instead of Celsius. \begin{align} Q &= mc\Delta T\\ \end{align} Now you can form the equations for each of the problem, while ...


3

There are multiple issues here, and the answer is: it depends. How much of your substance is there relative to the solution? If the mass is, say, 1% of total and the specific heat is comparable, then your answer may be off by 1%, which is probably smaller than the systematic measurement error from heat loss in the calorimeter. What is the specific heat of ...


3

First, we have some things to consider and they are: the specific heat of water; the latent heat of vaporization of water; how the terms mentioned above are compared to each other. The specific heat of water is $\pu{4200 J kg-1 K-1}$ and it means that it takes $\pu{4200 kJ}$ amount of heat to increase the temperature of $\pu{1 kg}$ water by $\pu{1 K}.$ On ...


3

The equations you listed can be used to partially solve the problem. However, with the equations you used, you have made one error which was in your choice of "moles." Your moles value must match up with your $\ce{C_{p}}$, so you must find the moles of Water in $\pu{95.00mL}$! Alternatively, you can use $4.184 \pu{\frac{J}{g\ ^\circ C}}$ and $\pu{1 \frac{...


2

Actually, a calorimeter with high specific heat would have less temperature change, which is harder to measure. Therefore, the error in the calculated enthalpy of the reaction would be higher. Specific heat does not mean heat conductivity, as described by Fourier's Law.


2

This is addressed experimentally, not through theory or computation. You run one experiment starting with the reactants, measuring the change in temperature over time. Then, you run a control with a similar solution that does not react (e.g. the product mixture), setting the mixer to the same speed. This will allow you to correct for any effect not due to ...


1

It models an isobaric process: The titration is performed at constant pressure and temperature, meaning that a single ITC experiment offers data on the binding enthalpy, the equilibrium association constant and the stoichiometry, from which the entropy of binding and Gibbs energy can be computed. Hence, a single ITC experiment offers direct access to the ...


1

Let's take a numerical example. $0.2432$ $g$ magnesium metal ($0.01$ mol) is introduced into a calorimeter containing $50~ g$ HCl $1$ M. The chemical reaction will be $$\ce{Mg + 2 H^+ -> Mg^{2+} + H2}$$ The composition of the solution is changing between the beginning and the end of the reaction. There is an excess of HCl. But let's admit in first ...


1

Answering your question as it stands is impossible, as we readers do not have enough information about your actual experiment. What do you mean with 'enthalpy of 1-butanol'? You can only measure changes in enthalpy, so you need to specifiy what 'change' you're looking at. You could mean the enthalpy of evaporation or the reaction enthalpy of some chemical ...


1

I don't think designers care about the specific heat of the materials. Their primary considerations are going to be standard engineering considerations, e.g. strength, lightness, cost, ease of fabrication, resistance to corrosion and wear. Aluminum scores well on all of these, so it's a very common material for use in building small devices. Styrofoam can ...


1

You have more or less answered your own question. Enthalpy is a function of state, so you can write an expression of the form $\Delta {H} = {H_f}-{H_i}$, and we know that this depends only on the initial and final states, and not the path taken between them. You also have the relation, $H = U + PV$, using this and the previous equation, and the fact that $U$...


1

Even though the pressure remains constant, heat can get in (or out) of the system. From the cup calorimeter, we are trying to find $\Delta H$ which is equal to the heat evolved from the system at constant pressure. We measure this heat and our job is done. $$\Delta H=q_p$$ If we want to calculate $\Delta U$, we just to find the heat evolved from the bomb ...


1

This graph comes from a so-called "differential thermal analysis". It is studying the effects of temperature on glass. Differential thermal analysis (DTA) is a method that heats two sample pans, one with a sample and one empty as a reference so that the effects of the sample pan can be negated. The heating is performed at a constant heat flow and the ...


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