Thermodynamics question (temp change, entropy, energy)...?
914713
Just assume the energy gained is the potential gravitational energy U=mgh, with h=100 meters, and then the heat capacity of water, etc.
800141
The question is if you start with gas at 20.0 atm and 20.3 C and heat it to 125 C what will be the pressure?  It is a changing condition problem making use of Charles' Law.  The answer is 27.1 atm.
741934
This question is not very difficult, but you have to be careful with the unit.\n\nFirst, to solve this question, you have to make the assumption that there is no energy lost to the surrounding. So, this reduces to a two system question: one system releases energy while the other one absorbs energy. In this particular case, the energy released by one system equals to the energy absorbs by another system.\n\nTo picture this, consider the following, one system goes through the following steps:\n\n[100 degC H2O (g)] --- (step 1) ---> [100 degC H2O (l)] --- (step 2) ----> [79 deg C H2O (l)]\n\nThe other system goes through:\n\n[260g, 19 degC H2O (l)] --- (step 3) ----> [260g, 79 degC H2O (l)]\n\nIt is obvious that both the energy involved in step 1 and 2 is negative (release energy) while the energy involved in step 3 is positive (absorb energy). We will now solve this question backwards. Starts from finding out the energy required by step 3, then find out the mass of steam that is needed to do the job.\n\nEnergy absorbed during step 3:\nE3 = m * c * (delta T) = 260 * 4.18 * (79-19) = 65208 J\n\nEnergy released in step 1:\n(you have to first convert heat of vaporization to unit of J/g: 40.7*1000/18 = 2261.111 J/g)\nE1 = m' * l = m' * 2261.111 = 2261.111m'\n(since we don't know how much steam is needed, let's just leave it as a variable)\n\nEnergy released in step 2:\n(the temperature of the same mass as above will be lowered to 79 degC)\nE2 = m' * 4.18 * (100-79) = 87.78m'\n\nTotal energy released = (2261.111 + 87.78)m' = 2348.89m'\n\nEnergy released = energy absorbed\n2348.89m' = 65208\nm' = 27.76 g
327565
Because there is no temperature change. Temperature is a physical measurement of the phenomena of heat transfer to and from a system, not of redistribution of energy within a closed system. The process you describe occurs within the closed system, and cannot be recorded as a release of heat energy from the system or intake of heat energy into the system.
1104550
Thermodynamics, field of physics that describes and correlates the physical properties of macroscopic systems of matter and energy. The principles of thermodynamics are of fundamental importance to all branches of science and engineering.\n\nA central concept of thermodynamics is that of the macroscopic system, defined as a geometrically isolable piece of matter in coexistence with an infinite, unperturbable environment. The state of a macroscopic system in equilibrium can be described in terms of such measurable properties as temperature, pressure, and volume, which are known as thermodynamic variables. Many other variables (such as density, specific heat, compressibility, and the coefficient of thermal expansion) can be identified and correlated, to produce a more complete description of an object and its relationship to its environment.\n\nWhen a macroscopic system moves from one state of equilibrium to another, a thermodynamic process is said to take place. Some processes are reversible and others are irreversible. The laws of thermodynamics, discovered in the 19th century through painstaking experiment, govern the nature of all thermodynamic processes and place limits on them.
857691
Molecules changing state from liquid to gas...entropy increasing...S is positive.
870952
To answer this Question you need to use the Conservation of Energy Law.   This takes its simplest form in the idea that:\n\nHeat Lost by the Steam    =   Heat Gained by the cold water\n\nYou will need to say:\nLet L be the unknown Quantity and form an Equation:\n\nHeat Lost by the Steam:\nThe steam (one assumes at 100Celsius) condenses to water at 100Celsius.  In doing this it gives up its Latent Heat of Vaporization (mL)   The water it forms is then cooled to the final temperature giving up this Internal Energy to the System.  The Internal Energy(Heat) gained is given by m.s.dT where dT is the change in temperature.\nNote that we have two  Energies to Add Together here.\n\n\nHeat Gained by the cold water:\nThe Internal Energy(Heat) gained is given by m.s.dt where dt is the change in temperature.\nBoth the cold water and the Calorimeter are heated from their starting temperature(15C) to the final temperature(70.6C)\nNote that we have two  Energies to Add Together here.\n\nNow write down the equation:\nHeat Lost by the Steam (steam + hot water)     =   Heat Gained by the cold water (cold water + calorimeter)\n              \nm1.L + m1.s.dT  =  m2.s2.dt + m3.s3.dt\nAnd solve for the Unknown L
817267
I think such questions bug you because you equate Internal energy with heat. Work and heat both are energies in transit for any process. Think like this, if 220 J work is done on the system its energy would have increased by 220 J.  But instead its internal energy has decreased by 500 J. So only way this is possible is if [220 + 500] J = 720 J amount of energy leaves the system as heat.  If you like to work with formula then\n\nU = Q - W\n\nhere U = -500; Q unknown: W = -220\n\n-500 = Q - (-220)\n-500 = Q + 220\n\nor Q = -500-220 =  -720 J\n\nSo the system has lost 720 J as heat
958888
Use the formula:\nQ (heat) = Cp (Heat Capacity) * delta T (Change in Temperature)
304984
I would need more information to help you answer that question.\n\nWhat is the mass of the metal being used? What was the original temperature of the metal?\n\nThe formula you need is:\n\nq = mc(Tf - Ti)\n\nq = energy change\nm = mass\nc = specific heat\nTf - Ti = change in temperature
20445
These are physical changes.  The matter simple changes state, depending on tempature.  Also see evaporation.
39424
heat is the measurement of thermal energy
811304
let thetemperature of the mix be 't'*Cfind the heat lost by coffee\nfind the heat gained by ice in three stages\na)from -2.2*C to 0*C\nb)for melting and becoming water at 0*C\nc)for getting to the equilibrium temperature of the mixture\nequate and solve
772410
electrical energy-------> heat + light\ndoes that help?
294164
Thw formular to use here is Q = mcT where Q = quantity of heat, m = mass, c = specific heat capacity, and T is the change in the temperature, thus here Q = 2.0g*(30-25)*0.90 = 9J. Notice that I sed Joules as the unit.\n\n2.For this one, a law which i cant remember the name right now is nedeed. Here we have C(s)+O2(g)=CO2 Enthalphy -390KJ\nand also Mn(s)+O2(g)=MnO2(s) Enthalphy = -520KJ\nFor this law to be appplied, we have to write the second equation the other way to become MnO2= Mn+O2 Enthalphy = 520KJ. Notice that i changed the sign when i reversed the equation. Then O2 could be canceled out on the left side of the first equation and on the right side of the second equation. This was we are adding. Finally adding, we get MnO2(s)+C(s) = Mn + CO2 Enthalphy = -390KJ+520KJ = 130KJ.\nand thats the correct answer
275638
- All of the above\n\nThe 1st Law of Thermodynamics tells us that energy is neither created nor destroyed, thus the energy of the universe is a constant. However, energy can certainly be transferred from one part of the universe to another. To work out thermodynamic problems we will need to isolate a certain portion of the universe (the system) from the remainder of the universe (the surroundings).
436257
i doubt your teacher couldnt answer these unless (s)he is not actually a physics teacher\n\nc and a\n\nthe initial momentum is 4kgm/s\nthe final momentum is conserved so it is still 4 kgm/s\nmomentum is mass*velocity\nmass is 1+3= 4kg\nv = p/m = 4kgm/s / 4kg = 1 m/s\n\nentropy is randomness or disorder, when the rooms are different temps, its like the temps are "sorted". there is order to the temperature. when they combine, there is no more order, the temperatures are more random instead of being focused around 2 different temps so disorder, or entropy increases