2. List three sources of error that could account for the differences between your values for the enthalpy of fusion of water and the accepted value of 334 J/g.

1 trial :  nothing is given for result comparision - so we have no idea if it's a mistake.

2nd trial : The results can be compared - if varies, one may go wrong, but which one?

3rd trial : If 3rd result is different from 1st and 2nd, it is unreliable.

calculating 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. assuming copper calorimeter , 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.

squared differences average (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 ignore them.

% error in L = same as in n = 7% x 547.4 = 40

The result is (i) L= 549 +/- 3 J/g or (ii) L = 550 +/- 40 J/g.

Both are very far above  334 J/g, so there is at least one systematic error  

e.g: 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 best.

However, the spread in the actual results is much smaller

* measurements were "fiddled" to get better results; other Trials were made but only best 3 were chosen.

L=(mc+MC)T/n is too high, so n (ice melted) may be too small, or T (temp fall) too high - why?

* we have assumed initial and final temperature of ice was 0c, it may actually have been colder, so less ice would melt -which 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.

common management practices in dairy and poultry farming are: (i) proper shelter facilities and their regular cleaning. (ii) some basic hygienic conditions such as clean water, nutritious food, etc. (iii) animals are kept in spacious, airy, and ventilated place.

if the hares very faster than they are now, and thus the lynx's hunting success rate drops because it will be harder to catch them, the rate that will change will be the pray-rate. the pray-rate will change because more hares will be able to survive, thus more individuals for reproduction, so the end result will be increasing population of hares.

the predator-rate will remain pretty much the same because there will be more than enough pray, but the hunting success rate is stagnant.