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Specific Heat Capacity and Latent Heat Questions - GCSE standard

Q10. Johnny investigated the thermal conductivity of different metals.

This is the method he used:

1. Measure the mass of an ice cube.

2. Put the ice cube on a metal block which is at room temperature.

3. Measure the mass of the ice cube after one minute.

4. Repeat with other blocks of the same mass made from different metals.

Here are Johnny's results:

Metal
Initial mass of ice cube /g
Final mass of ice cube /g

Change in mass of ice cube /g
Aluminium
25.85
21.14
4.71
Copper
26.20
20.27
5.93
Lead
25.53
21.97
3.56
Steel
24.95
19.45
5.50

 

(a) The initial temperature of each ice cube was –15 °C

Choose one of the following sentences, by ticking the box, to explain why was it important that the initial temperature of each ice cube was the same.

Initial temperature was a continuous variable.  
Initial temperature was a control variable.
Initial temperature was the dependent variable.  
Initial temperature was the independent variable.  

 

[1 mark]

(b) Which metal had the highest thermal conductivity? Give a reason for your answer.

Copper had the highest thermal conductivity as there was a greater change in the mass of the ice than with the other metals (or the ice melted faster).

[2 marks]

(c) Suggest one source of random error in the student's investigation.

Any one from:

a variation in initial mass of ice cube

variation in the surface area of the ice cube touching the metal

or melting of ice while handling

variation in room temperature

initial temperature of metal block

[1 mark]

(d) An ice cube has a temperature of –15.0 °C

The total thermal energy needed to raise the temperature of this ice cube to 0.0 °C and completely melt the ice cube is 5848 J

Given that:

specific heat capacity of ice = 2100 J/kg °C

specific latent heat of fusion of ice = 334 000 J/kg

Calculate the mass of the ice cube.

Energy to raise the temperature to 0.00 °C = mcΔT

Energy to melt the ice = mL

Total energy required = mcΔT + ml

Always explain what you are doing - so the examiner understands your working, and if you make a mistake with numbers you can still get some credit.

E = m (cΔT + L)

m = E/(cΔT + L)

m = 5848/(2,100 x 15 + 334,000)

m = 5848 / (31,500 + 334,000)

m = 5848 / 365,500

m = 0.016 kg

m = 16 g

[5 marks]

(9 marks total)