The glow emitted by a substance exposed to external radiation is called 'fluorescence'. In fluorescence, a fluorophore is exposed to external radiation, absorbs the energy and emits a form of light or glow. The answer to this question is 'fluorescence'. I hope this helps.
Explanation:
P1V1 = nRT1
P2V2 = nRT2
Divide one by the other:
P1V1/P2V2 = nRT1/nRT2
From which:
P1V1/P2V2 = T1/T2
(Or P1V1 = P2V2 under isothermal conditions)
Inverting and isolating T2 (final temp)
(P2V2/P1V1)T1 = T2 (Temp in K).
Now P1/P2 = 1
V1/V2 = 1/2
T1 = 273 K, the initial temp.
Therefore, inserting these values into above:
2 x 273 K = T2 = 546 K, or 273 C.
Thus, increasing the temperature to 273 C from 0C doubles its volume, assuming ideal gas behaviour. This result could have been inferred from the fact that the the volume vs temperature line above the boiling temperature of the gas would theoretically have passed through the origin (0 K) which means that a doubling of temperature at any temperature above the bp of the gas, doubles the volume.
From the ideal gas equation:
V = nRT/P or at constant pressure:
V = kT where the constant k = nR/P. Therefore, theoretically, at 0 K the volume is zero. Of course, in practice that would not happen since a very small percentage of the volume would be taken up by the solidified gas.
<span>1 trial : you have nothing to compare the result with - you don't know if it's a mistake.
2 trials : you can compare results - if very different, one may have gone wrong, but which one?
3 trials : if 2 results are close and 3rd far away, 3rd probably unreliable and can be rejected.
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First calculate the enthalpy of fusion. M, C and m,c = mass and
specific heat of calorimeter and water; n, L = mass and heat of fusion
of ice; T = temperature fall.
L = (mc+MC)T/n.
c=4.18 J/gK. I assume calorimeter was copper, so C=0.385 J/gK.
1. M = 409g, m = 45g. T = 22c, n = 14g
L = (45*4.18+409*0.385)*22/14 = 543.0 J/g.
2. M = 409g, m = 49g, T = 20c, n = 13g
L = (49*4.18+409*0.385)*20/13 = 557.4 J/g.
3. M = 409g, m = 54g, T = 20c, n = 14g
L = (54*4.18+409*0.385)*20/14 = 547.4 J/g.
(i) Estimate error in L from spread of 3 results.
Average L = 549.3 J/g.
average of squared differences (variance) = (6.236^2+8.095^2+1.859^2)/3 = 35.96
standard deviation = 5.9964
standard error = SD/(N-1) = 5.9964/2 = 3 J/g approx.
% error = 3/547 x 100% = 0.5%.
(ii) Estimate error in L from accuracy of measurements:
error in masses = +/-0.5g
error in T = +/-0.5c
For Trial 3
M = 409g, error = 0.5g
m = 463-409, error = sqrt(0.5^2+0.5^2) = 0.5*sqrt(2)
n =(516-463)-(448-409)=14, error = 0.5*sqrt(4) = 1.0g
K = (mc+MC)=383, error = sqrt[2*(0.5*4.18)^2+(0.5*0.385)^2] = 2.962
L = K*T/n
% errors are
K: 3/383 x 100% = 0.77
T: 0.5/20 x 100% = 2.5
n: 1.0/14 x 100% = 7.14
% errors in K and T are << error in n, so we can ignore them.
% error in L = same as in n = 7% x 547.4 = 40 (always round final error to 1 sig fig).
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The result is (i) L= 549 +/- 3 J/g or (ii) L = 550 +/- 40 J/g.
Both are very far above accepted figure of 334 J/g, so there is at least
one systematic error in the experiment or the calculations.
eg calorimeter may not be copper, so C is not 0.385 J/gK. (If it was
polystyrene, which absorbs/ transmits little heat, the effective value
of C would be very low, reducing L.)
Using +/- 40 is probably best (more cautious).
However, the spread in the actual results is much smaller; try to explain this discrepancy - eg
* measurements were "fiddled" to get better results; other Trials were made but only best 3 were chosen.
* measurements were more accurate than I assumed (eg masses to nearest 0.1g but rounded to 1g when written down).
Other sources of error:
L=(mc+MC)T/n is too high, so n (ice melted) may be too small, or T (temp fall) too high - why?
* it is suspicious that all final temperatures were 0c - was this
actually measured or just guessed? a higher final temp would reduce L.
* we have assumed initial and final temperature of ice was 0c, it may
actually have been colder, so less ice would melt - this could explain
small values of n
* some water might have been left in container when unmelted ice was
weighed (eg clinging to ice) - again this could explain small n;
* poor insulation - heat gained from surroundings, melting more ice,
increasing n - but this would reduce measured L below 334 J/g not
increase it.
* calorimeter still cold from last trial when next one started, not
given time to reach same temperature as water - this would reduce n.
Hope This Helps :)
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From the calculations, the heat that is required is 2.1 kJ.
<h3>What is the specific heat capacity?</h3>
The term specific heat capacity has to do with the amount of heat that must be supplied to 1Kg of a substance in order to raise its temperature by 1K.
In this case;
H = mcdT
H = 100 grams * 4.18 J/gC * (25 - 20)
H = 2.1 kJ
Learn more abut specific heat capacity:brainly.com/question/1747943
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