You heat 1 L of water in a kettle on a stove burner and notice that the water te
ID: 991038 • Letter: Y
Question
You heat 1 L of water in a kettle on a stove burner and notice that the water temperature increases with time according to the following relationship:
T = -0.2044t3+2.2075t2+5.4644t+22.598 where T is the temeprature in celsius and t is the time in minutes.
Use the transient energy balance equation to determine the values of Q for the different values of t below. Neglect any mass lost by evaportation.
I need Q for t = 0.32 min, 2.50 min, and the amount of heat added from t = 0 min to t = 5.6 minutes.
The hint says:
The transient energy balance equation for a closed system is M(Cv)(dT/dt) = Q - W
where M is the mass or moles of the system, Cv is the heat capacity of the system, dT/dt is the change of temperature with time, Q is the flow rate of heat into the system, and W is the rate of work. There is no work in this system, and M can be found using the density of water. Cv is found in your text book as 75.4 × 10–3. dT/dt can be found by taking the derivative of the time vs temperature expression in the problem statement.
(dT/dt) = -0.6132t2 + 4.415t + 5.4644 celsius/min
Please help with all 3 subparts.
Explanation / Answer
The energy balance equation can be written as M(Cv)(dT/dt) = Q +W
where M is the mass or moles of the system, Cv is the heat capacity of the system, dT/dt is the change of temperature with time, Q is the flow rate of heat into the system, and W is the rate of work done on the system. There is no work in this system.
There is no work and hence W=0
MCv*dT/dt= Q (1)
Cv=75.4*10-3 KJ/mole.deg.c
Volume of water =1 L. Assuming density of water to be 1 g/cc, mass of water= 1000 gm
=1000/18 gmoles=55.56 gmoles
T=-0.2044t3+2.2075t2+5.4644t+22.598
dT/dt= -3*0.2044t2+2*2.2075t+5.4644 =-0.6132t2+4.415t
Eq.1 now becomes
55.56*75.4*10-3* (-0.6132t2+4.415t)= Q
At t=0.32min, Q= 55.56*75.4/1000 { -0.6132*(0.32)2 +4.415*0.32)= 5.665 Kj/min
Ar t= 2.5 min, Q= 4.19{ (-0.6132*(2.5)2 +4.415*2.5)}=30.189 Kj/min
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