Question #191438



  1. A student carries out an experiment to find the specific heat capacity of a liquid.  The results from the experiment were as follows


Initial temperature = 190

Final temperature = 350C

Mass of liquid = 2kg

Power of heater = 350W

Time heater on = 5min

a)     Calculate the energy supplied by the heater

b)     Calculate the specific heat capacity of the liquid

c)     On reviewing the experiment the student found that the container of the liquid was a poor insulator of heat.  Would the calculated value of the specific heat capacity of the liquid be higher or lower than the actual value?


Expert's answer

(1)

Energy supplied

Q=powertimes=3505×60=1.17JQ=\frac{power}{times}=\frac{350}{5\times60}=1.17J


(2)

Specific heat

S=QmΔTS=\frac{Q}{m\Delta T}


S=1.17(2)(35−19)=0.036J/kg°CS=\frac{1.17}{(2)(35-19)}=0.036J/kg°C

(3) due to pure insulation some amount of heat is dissipated in surrounding so change in heat supplied is reduced so specific heat of liquid will be lower then actual value.




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