FOLLOW US ON TWITTER
SHARE THIS PAGE ON FACEBOOK, TWITTER, WHATSAPP ... USING THE BUTTONS ON THE LEFT


YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Saturday, March 2, 2019

A student carries out an experiment to determine the specific heat capacity of a liquid using the apparatus illustrated in Fig. 3.1.


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

4 comments:

  1. Why is the valuw of R in 2 sf ?

    ReplyDelete
    Replies
    1. resistances are usually given as whole numbers

      Delete
  2. Isnt it supposed to be P=(m/t)c(thelta) +h instead of P=mc(thelta)+h ?

    ReplyDelete
    Replies
    1. yes, but the question says that the mass per unit time is m.

      so, what we usually label as m/t is actually represented as 'm' in this question

      Delete

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | A student carries out an experiment to determine the specific heat capacity of a liquid using the apparatus illustrated in Fig. 3.1.