# Homework Help: Uncertainties help

1. Jun 19, 2010

### spoony33

I have this question for homework and i'm well stuck!

The specific heat capacity of a liquid was found by heating a measures quantity of the liquid for a certain length of time. The following results were obtained.
Power of heater: ( 50.0 ± 0.5)W
Mass of liquid: (200 ± 10)g
Time of heating: (600 ± 1)s
Temperature rise: (50.0 ± 0.5)°C

I managed to get the percentage error in each reading but i just don't know how to work out the approximate percentage error in the value of specific heat capacity!

A) What will the approximate percentage error in the value of specific heat capacity?
B) Suggest one way in which to reduce the percentage error obtained for the specific heat capacity?

I managed to work out this but i dont know how to work out the approximate percentage error in the value of specific heat capacity?
The percentage of uncertainty of each is:
Power of heater:
(±0.5)/50.0 = ± 1%

Mass of liquid
(±10)/200 = ± 5%

Time of heating:
(±1)/600 = ±0.167 %

Temperature Rise:
(±.5)/50.0 = ± 1%

Thanks

The equation of specific heat capacity is: Eh = mcΔT

E = Energy, J
m = mass, kg
c = Specific heat capacity, J kg-1 °C-1
ΔT = Change in temperature, °C

Last edited: Jun 19, 2010
2. Jun 19, 2010

### hikaru1221

So C = (Eh)/(mΔT), right?
If a = b*c, what is the relation of Δa/a, Δb/b, Δc/c? (Δa/a is the percentage error of a).

Last edited: Jun 19, 2010