The specific heat of water (c_w) is larger than the specific heat of sand (c_s),
ID: 1998285 • Letter: T
Question
The specific heat of water (c_w) is larger than the specific heat of sand (c_s), i.e., cw > c_s. Explain how you derive the flow of wind (from the beach to the ocean, or from the ocean to the beach) along the coastline of an island during day time and during night time. Consider the following two-step process shown in the figure. Heat is allowed to flow out of an ideal gas at constant volume so that its pressure drops from 2.2 atm to 1.4 atm. Then the gas expands at constant pressure, from a volume of 6.8 L to 9.3 L, where the temperature reaches its original value. Find (i) the total work done by the gas in the whole process. (ii) the change in internal energy of the gas in the whole process. (iii) the total heat flow into or out of the gas (you are to compute the total heat flow in the whole process and specify whether the heat flows into or out of the gas).Explanation / Answer
Number of moles n = 1 mol ( say)
Pressure at A is PA = 2.2 atm = 2.2 x1.01 x10 5 Pa
Volume at point A is VA = 6.8 L = 6.8 x10 -3 m 3
Tempearture at point A is TA = PAVA /nR
=( 2.2 x1.01 x10 5 )(6.8 x10 -3)/(1)(8.314)
= 181.73 K
Temperature at point B is TB = ?
At constant volume PA/PB = TA/TB
From this TB = TA(PB/PA)
= 181.73 ( 1.4 atm / 2.2 atm )
= 115.65 K
Temperature at point C is TC = ?
At constant pressure , TC/TB = VC/VB
TC = TB(VC/VB)
= 115.65 (9.3 /6.8)
= 158.16 K
From point a to point B :
Work done Wab = 0
Since the change in volume in this process is zero.
Heat Qab = Cv(TB-TA)
Where Cv = Specific heat at constant volume = 2.5 R ( let the gas is diatomic gas)
= 2.5 x 8.314 J / mol K
Substitute values you get ,
Qab = 2.5 x 8.314 x ( 115.65 - 181.73)
= -1373.4 J
Change in internal energy Uab = Qab = -1373.4 J
Process from point b to point c :
Work done W bc = PB ( VC-VB)
= 1.4 atm (9.3 L - 6.8 L)
= 1.4 x1.01x10 5 x2.5 x10 -3 m 3 SInce 1 L = 10 -3 m 3
= 353.5 J
Heat Qbc = Cp(TC-TB) Where Cp = Specific heat at constant pressure = (7/2) R = 3.5 R
= 3.5(8.314) ( 158.16 - 115.65 )
= 1237 J
Change in internal energy Ubc = Qbc - Wbc
= 883.5 J
From process from point c to point a :
It is adiabatic process :
Heat Q ca = 0
Work done Wca = [-R/(r-1) ] [Ta-Tc]
Where r = ratio of specific heats = 1.4
Substitute values you get , Wca = [-8.314/(1.4-1)][181.73 - 158.16]
= -490 J
Change in internal energy Uca = - Wca = 490 J
(a). Total work done W = Wab + W bc + W ca
(b).Total change in internal energy U = Uab + Ubc+Uca
(c).Total heat Q = Qab + Qbc+Qca
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