Question

A student carries out an experiment to find the specific heat capacity of aluminium. He uses
an electric heater and a thermometer inserted into separate holes in an aluminium block.
The following data are obtained.
mass of aluminium block = 2.0 kg
power of heating element = 420 W
time of heating = 95 s
initial temperature of block = 19.5 °C
final temperature of block = 40.5 °C
Calculate the value of the specific heat capacity of aluminium given by this experiment.
(c) In the experiment in (b), no attempt is made to prevent loss of thermal energy from the surfaces of the block.
Suggest two actions the student could take to reduce the loss of thermal energy from the surfaces of the block.

Answer 1

Answer:

Explanation:

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           

Hope it helps...

Answer 2

(b) Answer:  950J/Kg℃

Explanation: Energy= Power x Time

Energy= 420x95= 39900

C= H/m △t

C= 39900/ 2Kg x (40.5-19.5)

C= 3990/42

C= 950J/Kg℃