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 V_rms = V₀ / √2 = 27 / √2 = 19.1 V}

V₀ = 27                                    or V_rms = 19.1

{Since the voltage changes with time, we need to use the rms voltage in calculations.

P = (V₀/√2)² / R                       or P = V_rms² / R

P = V₀² / 2R                            or P = V_rms² / R }

33.3 = 27² / 2R                        or 33.3 = 19.1² / R

R = 11 Ω