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?

Answer 1

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