Question 3
(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]
Reference: Past Exam Paper – March 2016 Paper 42 Q3
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
Why is the valuw of R in 2 sf ?
ReplyDeleteresistances are usually given as whole numbers
DeleteIsnt it supposed to be P=(m/t)c(thelta) +h instead of P=mc(thelta)+h ?
ReplyDeleteyes, but the question says that the mass per unit time is m.
Deleteso, what we usually label as m/t is actually represented as 'm' in this question