Alternative B
Question 2
- You are provided with a bunsen burner, weighing balance, a thermometer, glass
stirrer, pyrex beaker, metal block, lagged calorimeter, measuring cylinder, water
and other necessary apparatus.
- Weigh and record the mass, Mc, of the calorimeter.
- Measure and record the temperature, θR, of the laboratory.
- Use the measuring cylinder to transfer a volume V = 50 cm3 of water into the calorimeter.
- Calculate the mass Mw of water transferred, given that the density of water is 1000 kg m-3.
- Clamp the thermometer as shown in the diagram above.
- Place the metal block in the beaker and add a reasonable amount of water to cover the block completely.
- Heat the mixture and allow it to boil for about 3 minutes.
- Transfer the hot metal block quickly into the water in the calorimeter and stir.
- Read and record the highest temperature, θ, attained by the mixture.
- Evaluate the rise in temperature T = (θ – θR) and T-1.
- Repeat the procedure for four other volumes V = 70 cm3, 90 cm3, 110cm3 and 130 cm3. In each case, record Mw and θ and evaluate T and T-1.
- Tabulate the results.
- Plot a graph of Mw on the vertical axis and T-1 on the horizontal axis.
- Determine the slope, s, of the graph and the intercept, c, on the vertical axis.
- Calculate k from the equation c = -(McK/4200)
- State two precautions taken to ensure accurate results.
-
- State the low of conservation of energy.
- A metal block of mass 0.50 kg is heated to 100oC and quickly immersed in 0.45 kg of water at 20oC. Calculate the equilibrium temperature of the mixture. [Specific heat capacity of water = 4200Jkg-1K-1;Specific heat capacity of metal = 134Jkg-1K-1].
Observation
Part (a) Candidates performance was poor. Many candidates did not record V in cm3 on the table,
hence lost mark for composite table. Results were not recorded in appropriate d.p. and s,f.
Part (b) Candidates response was unsatisfactory.
The expected response:
3. (a) OBSERVATIONS [07]
(i) Mc recorded in g or kg.
(ii) θR, recorded in °C
(iii) Five values of V recorded in cm3
(Deduct ½ mark for each wrong or missing value)
(iv) Five values of Mw correctly evaluated and recorded to at least 2 d.p. in g or kg.
(Deduct ½ mark for each wrong or missing value)
(v) Five values of θ recorded in °C and in trend.
Trend: (As V increases, θ decreases)
NOTE: θ ˃ θR
(Deduct ½ mark for each wrong or missing value)
(vi) Five values of T = (θ – θR) correctly evaluated.
(Deduct ½ mark for each wrong or missing value)
(vii) Five values of T-1 correctly evaluated to at least 3 d.p.
(Deduct ½ mark for each wrong or missing value)
(viii) Composite table showing at least V, Mw, T, T-1 and θ.
GRAPH [06]
(i) Both axes correctly distinguished (½ mark each)
(ii) Reasonable scales (½ mark each)
(iii) Five points correctly plotted
(Deduct 1 mark for each wrong or missing point)
(iv) Line of best fit
NOTE: Deduct 1 mark for d.i. if axes do not start from origin (0,0)
SLOPE [02]
(i) Large right-angled triangle
(ii) ∆Mw correctly determined
(iii) ∆T-1 correctly determined
(iv) ∆Mw/(∆T-1 ) correctly evaluated
STRAIGHT LINE [01]
Straight line making 45° with horizontal axis and intersecting the curve.
INTERCEPT [01]
Intercept, c, on the vertical axis
Correctly shown
Correctly recorded
CALCULATION [01]
Calculation of k
|k| = 4200c/Mc
Correct substitution
Correct evaluation
NOTE: If mass is in g, calculate k = 4.2c/Mc
ACCURACY [01]
Based on k = specific heat capacity of the calorimeter ± 10%
Copper = 400Jkg-1K-1
Aluminium = 900Jkg-1K-1
PRECAUTIONS [02]
Award 1 mark each for any 2 correct precautions stated in acceptable tense.
e.g
- Repeated readings (shown on table)
- Avoided parallax error in weighing balance
- Noted/Corrected zero error on weighing balance
- Thermometer did not touch the calorimeter
- Gentle stirring
- Avoided splashing of water
(b)(i) Energy is neither created nor destroyed but can be changed
from one form to another.
OR
In an isolated or closed system, the total energy is constant.
(ii) Let the equilibrium temperature be T
Heat lost by metal = heat gained by water
0.50 x 134(100 – T) = 0.45 x 4200(T – 20)
44500 = 1957T
T = 22.7°