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Physics

An electric heater of power 780 W raises the temperature of 5.5 kg of a lab-chemical from 30°C to 41°C in 1.5 min.

Calculate

(a) energy supplied by heater,

(b) heat capacity and

(c) the specific heat capacity of the lab-chemical.

Current Electricity

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Answer

(a) Given,

t = 1.5 min

converting min to sec

1 min = 60 sec

so 1.5 min = 60 x 1.5 = 90 sec

Rise in temperature = 41 - 30 = 11°C

Mass of liquid = 5.5 kg

Energy supplied by heater (E) = P x t

E = 780 x 90 = 70200 J = 70.2 kJ

(b) From relation, Heat capacity (C)

C = E△t\dfrac{E}{△t}

Substituting we get,

C = 7020011\dfrac{70200}{11} = 6381.8 J

(c) From formula, Specific heat capacity (s)

s = Em×△t\dfrac{E}{m \times △t}

Substituting we get,

s = 702005.5×11\dfrac{70200}{5.5 \times 11} = 1160.3 J

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