Energy supplied by heater = energy absorbed by aluminium Pt = mc T 200 240 = 1 c (70 20) c = 960 J kg 1 C 1 The specific heat capacity of aluminium is 960 J kg 1 C 1 . 1M 1M 1A (iii) Some energy is lost to the surroundings. The actual amount of energy absorbed by the metal block is less than the energy supplied by the heater. The answer in (ii) is therefore higher than the standard value. 1A 1A 5 Solutions Marks (a) Put the thermometer in boiling water and mark the mercury level. Then put the thermometer in melting ice and mark the mercury level. The temperatures of the boiling water and melting ice are taken as 0 C and 100 C respectively. The separation between these two mercury levels is divided into 100 equal divisions and each division is 1 C. 1A 1A 1A 1A (b) By proportion, room temperature = 0 . 4 0 . 24 0 . 4 4 . 8 100 1M = 22 C 1A (c) (i) Initial temperature = 22 C Final temperature = 0 . 4 0 . 24 0 . 4 5 . 15 100 = 57.5 C 1M Energy supplied by heater = energy absorbed by liquid L Pt = mc T 50 5 60 = 0.5 c (57.5 22) c = 845 J kg 1 C 1 1M 1M 1A The specific heat capacity of liquid L is 845 J kg 1 C 1 . (ii) The result of experiment is different from the actual value because energy is lost to the surroundings. 1A 5
LQ_Heat and Internal Energy Methods to reduce heat loss to the surroundings (any two of
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