Q#3 (Past Exam Paper – March 2016 Paper 42 Q3)
(a) Define specific heat capacity. [2]
(b) A student carries out an experiment to determine the specific heat capacity of a liquid using the apparatus illustrated in Fig. 3.1.
Fig. 3.1
Liquid enters the tube at a constant temperature of 19.5 °C and leaves the tube at
a temperature of 25.5 °C. The mass of liquid flowing through the tube per unit time is m.
Electrical power P is dissipated in the heating coil.
The student changes m and adjusts P until the final temperature of the liquid leaving the tube is 25.5 °C.
The data shown in Fig. 3.2 are obtained.
m / g s-1 P/ W
1.11 33.3
1.58 44.9
Fig. 3.2
(i) Suggest why the student obtains data for two values of m, rather than for one value. [1]
(ii) Calculate the specific heat capacity of the liquid. [3]
(c) When the heating coil in (b) dissipates 33.3 W of power, the potential difference V across the coil is given by the expression
V = 27.0 sin (395t ).
The potential difference is measured in volts and the time t is measured in seconds.
Determine the resistance of the coil. [3]
[Total: 9]
Solution:
(a) Specific heat capacity is the (thermal) energy per unit mass required to raise the temperature of a substance by one degree
(b)
(i) To allow for the heat losses to (or gained from) the surroundings.
(ii)
{Heat energy = mass × specific heat capacity × change in temperature
H = mcΔθ
Divide by time,
H/t = mcΔθ / t
Power = (m/t)cΔθ
where H/t is the energy / time which is the power
m/t is the mass of liquids flowing through the tube per unit time (in this question it is m – see the unit from the table; it is ‘g s-1’ and not ‘g’.)
To account for the heat exchange with the surroundings, we include ±h.}
EITHER P = mcΔθ ± h
{We can obtain 2 different equations from the 2 set of readings.}
OR 44.9 = 1.58×10–3 × c × (25.5 – 19.5) ± h
OR 33.3 = 1.11×10–3 × c × (25.5 – 19.5) ± h
{The value of h is actually unknown. BUT if we consider both equation at the same time, it can be eliminated by subtracting the 2 equations.}
(44.9 – 33.3) = (1.58 – 1.11) × 10–3 × c × (25.5 – 19.5)
c = 4100 (4110) J kg–1 K–1
(c)
{Power dissipated, P = V2 / R
The formula is usually of the form: V = V0 sin (ωt) where V0 is the peak voltage (from a.c.)
So, V0 = 27 V or Vrms = V0 / √2 = 27 / √2 = 19.1 V}
V0 = 27 or Vrms = 19.1
{Since the voltage changes with time, we need to use the rms voltage in calculations.
P = (V0/√2)2 / R or P = Vrms2 / R
P = V02 / 2R or P = Vrms2 / R }
33.3 = 272 / 2R or 33.3 = 19.12 / R
R = 11 Ω
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