1)A 55.67 g sample of a substance is initially at 22.7 °C. After absorbing 2847
ID: 891918 • Letter: 1
Question
1)A 55.67 g sample of a substance is initially at 22.7 °C. After absorbing 2847 J of heat, the temperature of the substance is 132.2 °C. What is the specific heat (c) of the substance?
2)An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its internal energy increased by 7785 J. What is the specific heat of the gas?
3)How much work must be done on a system to decrease its volume from 13.0 L to 5.0 L by exerting a constant pressure of 4.0 atm?
4)If a system has 5.00 × 102 kcal of work done to it, and releases 5.00 × 102 kJ of heat into its surroundings, what is the change in internal energy of the system?
Explanation / Answer
Q = m c T
Q = quantity of heat in joules (J)
m = mass of the substance acting as the environment in
grams (g)
c = specific heat capacity (4.19 for H2O) in J/(g oC)
T = change in temperature = Tfinal - Tinitial in oC
specific heat (c) = Heat Energy / (mass of substance * change in temperature)
specific heat (c) = 2847 / 55.67 x 109.5 = 0.4670 joules/ gram X degree Celsius
First you need to calculate q
delta U is change in internal energy
delta U = q + w
q is heat and w work done
here work was done by the system means energy leaving the system so w is negative
delta U = q + w
q = delta U - w = 7785J - (-346 J) = 8131 J = 8.131 KJ
q = m x c x delta T
8131 J = 80.0 g x c x (225-25) °C
c = 0.5081 J /g °C
Answer: W= p*(V1-V2)
you must use MKSA units 1L= 0.001m3
4.0atm =4.0*101.325Pa
W= (0.013-0.005) *4*101325=3242.4 J
5.00x102 kcal x 4.19 kj/kcal = 2.095 x 103 kj
Change in internal energy = 2.095 x 103 kj - 5.00x102kj = +15.95 x 102 kj
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