an iron block is heated using electric heater as shown?

Question

the joulemeter is measure the electrical energy supplied.when 7200j of electrical energy is supplied to heater the temperature of the 0.50kg block rises from 10 degree celious to 40 degree celious.
(a) calculate a value for the specific heat of capacity of iron.
(b) the actual value for the specific heat capacity of iron is 450j/(kgk) give a reason for the difference between these two values.
(c) state one way of improving the accuracy of the experiment.

Answer 1

Shown where?

(a)
Q= Cp m (T2-T1)
Q= 7200 Joules
Cp=?
T2= 40
T1= 10

Cp= Q/(m(T2-T1))
Cp=7200/(0.50 (40 - 10))
Cp=480 Joules/(kg K) Difference is the same between deg K or deg C .
(b) Heat losses.
(c) Better insulation, careful measurements.

Answer 2

a) Q = m c deltaT

re-arrange for c -:

c = Q / m deltaT

c = 7200 / (0.5 x (313 - 283))

c = 480 j/kg

b) The heater is not 100% efficient, and the block is losing heat constantly to its colder surroundings, meaning more energy from the heater was required to maintain the temperature.

c) Repeat the experiment at least 3 times and then take an average value from the 3 results, or insulate the block somehow so it doesn't radiate away its heat.

Answer 3

Heat = mass * specific heat capacity * change in temperature.
Heat is measured in Joules.
Mass in kg
Temperature in celcius.

Rearrange to find the specific heat capacity.

part c gives a clue as to why the value is different...it's innacurate. You could increase the mass of the iron to reduce the error.