(1) Let 5 million calories of solar energy be absorbed by 2 cubic meters of hydrogen gas 100 km above the Earth. If the particle density of the gas is 10,000 atoms per cubic meter, estimate the heat capacity of the gas volume. (Atomic mass of a hydrogen atom, 1 u ~ 1.6 x 10^-27 kg).
(2) 10 lbs. of iron and 5 lbs. of aluminum - both at 200 F, are added to 10 lb. of water at 40 F contained in a vessel whose thermal capacity is 0.5 Btu/ deg F. Calculate the final temperature if c(Al) = 0.21 cal/ g C, and c (Fe) = 0.11 cal/g C (Note: specific heat capacities are the same in calories per gram per degree Celsius as in Btu/ F deg).
(3)A calorimeter and its contents have a total thermal (heat) capacity of C = 200 cal/ deg C. A body of mass 210 g and at temperature 80 C is placed in the calorimeter resulting in a temperature increase from 10 C to 20 C. Compute the specific heat capacity of the body.
(1): The heat capacity Q = mc so: c= Q/m
Where Q = 5 x 10^6 cal
But J = W/H = 4.18 J per calorie, so:
Q = 4.18 (J/cal) x (5 x 10^6 cal ) = 2.1 x 10^7 J
The mass m (of hydrogen) in the volume is:
m = 2 m^3 x (10^5/m^3) x ( 1.6 x 10^-27 kg) = 3.2 x 10^-22 kg
Then the heat capacity = (2.1 x 10^7 J)/ (3.2 x 10^-22 kg) = 6.5 x 10^28 J/kg
(2) We use: Q lost by metals = Q gained by water + Q gained by vessel
Q lost by metals = m(Fe)c(Fe)(200F – T ) + m(Al)c(Al)(200F – T)
Q gained by water + vessel = (10 lb.)(T – 40F)
(10)(0.11)(200 – F) + (5)(0.21)(200 – F) = 10(T – 40F)
1.1(200- T) + 1.1(200 – T) = 10T – 400
440 – 2.2T = 10T – 400
12.2T = 840
T = 68F
(3) Cm (t2 – t1) = 200 cal/ deg C
(200 cal/C)(20 C – 10C) =210 g(c) (80 C – 20 C) = 2,000 cal
Then: c = (2,000 cal)/ (210)(60) = (2,000 cal)/ 12400 = 0.16 cal/g deg C
1) A student performs the heat conductivity experiment as shown in Fig. 1, and determines the thermal conductivity of copper to be 390 W/mC. If he then measures the thermometer differences (t4 - t3) = 5 C and (t2 - t1) = 2C, using 0.5kg of warmed water, and his copper test rod for the experiment is 0.5 m long, what would its cross-sectional area A have to be? (Take the specific heat capacity of water = 4200 J/kg K). Also, obtain the % of error in the student's result by looking up the actual thermal conductivity of copper.
2)A plate of copper 0.4 cm thick has a temperature difference of 60 C between its faces. Find: a) the temperature gradient, and b) the quantity of heat that flows through each square centimeter of one face each minute?
3)How many calories per minute will be conducted through a window glass 80 cm x 100 cm by 2mm thick if the difference between the two sides is 20C?
4) A group of 4 astronauts lands on Mars with solar radiation collection material of total area 2000 m^2. If the efficiency of the material is 30%, and the ambient night time temperature on Mars for their base location (Isidis Planitia) is -40 C (10C day time), will they have adequate collecting material if the solar constant on Mars is 620 W/m^2? (Assume insulating material with a thermal conductivity of 0.08 W/mC, and a need to keep the inside area of their domecile at least at 10 C, requiring solar radiant energy collected of at least 1,200 W per minute for an area of 10 m x 10 m.) Estimate the thickness of insulating material they're likely to need in order to make it work. Comment on whether this expedition is even feasible given the limits of their materials, and that no more than 100 m^3 of insulating material can be taken.
(1) We require:
k A T(t2 - t1)/L = mc(t4 - t3)
where: k = 390 W/mC , L = 0.5m, mass of water = 0.5 kg, T = 600 s
(t4 - t3) = 5 C and (t2 - t1) = 2C
Then: A = mc(t4 - t3) / k T(t2 - t1)/L = (1050 J/C)/ (1560 W/m^2 C)
[(0.5 kg)( 4200 J/kg C)(5 C)] / [(390 W/mC) (2C)/ 0.5m] = 0.67 m^2
The % error: Actual k = 397 W/mC so: % Error =
(397 – 390)W/mC/ 397 W/mC x 100% = 1.7%
(Note: the student likely has committed other serious errors even though he got close to the correct k. The reason is that the value for A is certainly excessive!)
(2) 0.4 cm = 0.004 m = x and (T2 – T1) = 60 C
(a) Temp. gradient = (T2 – T1)/x = 60 C/ 0.004 m = 1.5 x 10^4 C/ m
(b) Q = k(T2 – T1)/x = (397 W/mC)(60 C)/ 0.004m = 5.9 x 10^6 W/m^2
That is: 5.9 million J/s per square meter
Now, 1 cm^2 = (0.01 m)^2 = 10^-4 m^2
Then: Q(/cm^2) = (5.9 x 10^6 W/m^2) x(10^-4 m^2/cm^2)= 590 W/cm^2
Or 590 J per cm^2 per sec
Then over 1 minute: (60s)
= (590 J per cm^2 /sec)(60 sec) = 35, 400 J
(3)Heat conductivity for glass = 0.0025 cal/cm C
And x = 2mm = 0.2 cm
A = 80 cm x 100 cm = 8,000 cm^2
Then: k A T(t2 - t1)/L =
[(0.0025 cal/cm C) (8 x 10^3 cm^2)(60s) (20 C)]/0.2 cm
= 120,000 calories
(4)The astronauts require 1200 W/m2 for a living area of 10m x 10m, but this assumes adequate insulation, with k = 0.08 W/mC
The heat lost per minute (Q/t) over the domecile area (100 m^2) given a Martian exterior temperature of (-40 C) is:
1200 W/m^2 (60s)(100m^2) = 7,200,000 J for (t2 – t1) = (10C – (-40C)) = 50 C
The heat energy that can actually be potentially delivered will be the Martian solar constant x the total area of the solar collector x the efficiency:
Q(eff) = (620 W/m^2) (2000 m^2) (0.30) = 372,000 W = 372,000 J/s
and over 1 minute: 372,000 J/s(60s) = 2.2 x 10^7 J
However, only the fraction able to cover the net heat losses should actually be used (in addition one must store the rest on account of likely meteorite damage!) If 650m^2 of collecting material is used, then the energy delivered per minute will be:
7.26 x 10^6 J
or just above the heat losses of 7.2 x 10^6 J
But all this assumes the insulation material is adequate and the limit is that only 100 m^3 can be brought.
Then compute the ratio A/L = (7.2 x 10^6 J)/ (0.08 W/mC)(60s) 50 C = 3 x 10^4 m
This means that there simply isn't enough insulation material to do the job. Even if L = 1 m (i.e. 1 m or 3.3' thickness of insulation was used) it would have to cover an area of 3 x 10^4 m^2 (e.g. A/L = (3.0 x 10^ 4 m^2/ 1m = 3 x 10^4 m). But the putative area for the living space is only 100m^2 (10m x 10m) meaning the height of each wall would need to be: h = (3.0 x 10^4 m^2/ 10 m)= 3.0 x 10^3 m = 9,900' or 990 stories high!