Calculation block, heat capacity

On still another day, she is again disheartened to discover that the left-hand side of (S2.13-3) gives only 36. However, she notices that Junior has been playing with matches, and she observes that the room temperature is slightly warmer than usual So she carefully measures the heat of combustion of a test block and calculates (from the known heat capacity of the nursery contents) the expected room temperature change,... 

Near room temperature, the specific heat capacity of benzene is 1.05 J-(°C) 1-g 1. Calculate the heat needed to raise the temperature of 50.0 g of benzene from 25.3°C to 37.2°C. (b) A 1.0-kg block of aluminum is supplied with 490 kj of heat. What is the temperature change of the aluminum The specific heat capacity of aluminum is 0.90 J-(°C) 1-g l. 

Calculate the mass loss or gain for each of the following processes (a) a 50.0-g block of iron (specific heat capacity, 0.45 J-(°C)"1-g 1) cools from 600°C to... 

A substance, metallic in nature, is to be identified, and heat capacity is one of the clues to its identity. A block of the metal weighing 150g required 38.5 cal to raise its temperature from 22.8°C to 26.4°C. Calculate the specific heat capacity of the metal and determine if it is the correct alloy, which has a specific heat capacity of 0.0713 cal/g K. 

One mole of metal block at 1,000K is placed in a hot reservoir at 1,200K. The metal block eventually attains the temperature of the reservoir. Calculate the total entropy change of both the system (the metal block) and the surroundings (the reservoir). The heat capacity of die metal is given as... 

Figure 2, A, represents the experimental heat capacity data in the temperature range between 20° and 360° K. for H2 in Pd4H2—i.e., Pd4H2 minus the heat capacity of the palladium atoms in palladium black (7) and block palladium (5). In Figure 2, B, C, and D represent the similarly calculated experimental contributions for H2 in the other samples studied which had H/Pd ratios of 0.75, 0.25, and 0.125. Above 120° K. the results for palladium black are noticeably different from all of the others. This is apparently due to the fact that in palladium black, owing to the smallness of the particles, the lattice is somewhat more mobile. In Figure 3 all the experimental contributions of two hydrogen atoms to the heat capacity for alloys of compositions H/Pd = 0.75, 0.50, 0.25, and 0.125 are plotted between 35° and 85° K. (5). All the points lie on a single curve, within experimental error. Such a situation is difficult to conceive unless the hydrogens are similarly located with respect to each other in all samples.

Styrofoam cups and the heat necessary to raise the temperature of the inner wall of the apparatus. The heat capacity of the calorimeter is the amount of heat necessary to raise the temperature of the apparatus (the cups and the stopper) by 1 K. Calculate the heat capacity of the calorimeter in J/K. (d) What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter ... 

A coffee-cup calorimeter of the type shown in Figure 5.18 contains 150.0 g of water at 25.1 °C. A 121.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g-K. The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1 C. (a) Determine the amormt of heat, in J, lost by the copper block, (b) Determine the amount of heat gained by the water. The specific heat of water is 4.18 J/g-K. (c) The difference between your answers for (a) and (b) is due to heat loss through the Styrofoam cups and the heat necessary to raise the temperature of the inner wall of the apparatus. The heat capacity of the calorimeter is the amount of heat necessary to raise the temperature of the apparatus (the cups and the stopper) by 1 K. Calculate the heat capacity of the calorimeter in J/K. (d) What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter ... 

Suppose you have two blocks of copper, each of heat capacity Cv = 200.0 J K . Initially one block has a uniform temperature of 300.00 K and the other 310.00 K. Calculate the entropy change that occurs when you place the two blocks in thermal contact with one another and surround them with perfect thermal insulation. Is the sign of AS consistent with the second law (Assume the process occurs at constant volume.)... 

The next stage in the mathematical treatment is summarized in Fig. 4.17. It involves the measurement of a heat capacity which changes with temperature, but sufficiently slowly that a steady-state is maintained. The changing heat capacity causes, however, different heating rates for the sample and reference, so that the simple calculation of Fig. 4.16 caimot be applied. Equations (9) and (10) express the temperature difference between the block and sample, and block and reference, respectively. They are derived simply from Eqs. (1) and (2) of Fig. 4.16, with the overall heat capacity of the sample and calorimeter being equal to Cp + mcp, where Cp is the heat capacity of the empty calorimeter (water value of the empty sample holder), m is the sample mass and Cp is the specific heat capacity of the sample. 

The bottom sketch in Fig. 5.2 represents a drop calorimeter. As in the liquid calorimeter, the mode of operation is isoperibol. The surroundings are at (almost) constant temperature and are linked to the sample via a controlled heat leak. The recipient is chosen to be a solid block of metal. Because it uses no liquid, the calorimeter is called an aneroid calorimeter. The use of the solid recipient eliminates losses due to evaporation and stirring, but causes a less uniform temperature distribution and necessitates a longer time to reach steady state. The sample is heated to a constant temperature in a thermostat (not shown) above the calorimeter and then dropped into the calorimeter, where the heat is exchanged. The temperature rise of the block is used to calculate the average heat capacity. 

The most thorough treatment of uranium and plutonium aquo-ion equilibria over extended temperatures is that of Lemire and Tremaine [71]. This paper uses the systematic relationships developed by Criss and Cobble [72], which relate aquo-ion entropies, heat capacities, and their high-temperature behavior. Although the experimental determination of aquo-ion heat capacities has been dramatically advanced by the development of flow microcalorimeters [73,74], the only measurements of f-block aquo-ion heat capacities were made before this innovation [75,76]. Therefore, Lemire and Tremaine had to rely on estimated heat capacities for almost all of their calculations, and most of their equilibrium constants are uncertain by two or more orders of magnitude. Lemire [77] has also written a report on neptunium aquo-ion equilibria over extended temperatures. 

I Assume two adiabatic systems (a) and (b) each consisting of a steel block with the mass m = f.OOO kg. The mass-specific heat capacity of the steel is Cp = 0.481 kJ/kg K. In the initial state, the temperatures of the two blocks are 0a = 0°C and Of, = 100°C, respectively. At the time ti, the two steel blocks are brought into mutual thermal contact. At the time t2, the two blocks have reached thermal equilibrium at 0ab = 50°C. Calculate for the described process (1) — (2), the increase in entropy for the two subsystems ASa and ASf, respectively, as well as the increase in entropy AS at of the total system ... 

An adiabatic system contains two, thermally separated metal blocks A and B. Both of the blocks have the mass 1000 g and the specific heat capacity of the metal is Cp = 0.38 J/g K. In its initial state, block A is in thermal equihbrium at 0 °C and block B is in thermal equilibrium at 100 °C. The blocks are brought into thermal contact and after some time they have obtained equilibrium at the mutual temperature 50 °C. Calculate the total entropy increase AS. niv in the thermodynamic universe by this process and evaluate the result based on the Clausius inequality I... 

The PRF method has been used - to measure the ultrasound absorption coefficient of tissue, the value of which is difficult to obtain using invasive temperature measurements. The tissue sample was placed in a block of agarose gel and heated by ultrasound. The energy absorbed by the tissue diffuses slowly into the surrounding gel. Knowing the thermal capacity of the gel, and the temperature distribution measured by MRI, an estimate of the total absorbed energy can be made. The ultrasound pressure absorption coefficient (a) can be calculated as ... 

Dulong block To calculate performance- and capacity-related parameters, the lower heating value (LHV) is assessed according to Dulong s equation using the ultimate analysis, including the fuel moisture (as-received basis). 

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