Hello!
Here is the solution to your question!
Amount of heat as input = 65.3 W 124 s = 8097.2 J
Q = mc
so, c = Q/m
= 7800 J / (0.78 kg )
= 2506 J/kg/c
if some of the energy was used to heat the container , or the surroundings, then less energy would go into heating the liquid; if less energy caused the 3.99C temp increase, the specific heat must be less, so the value we have calculated is an overestimate.
For you to better understand this, let us sum up once again,
Energy supplied = 120*65 Ws = 8097 joules.
Energy supplied per kg = 7800/0.78 = 10000 joules/kg
rise in temperature = 22.54 - 18.55 C = 3.99C
10000 joules raised 1 kg by 3.99C
so 2506 joules raises 1kg by 1C
specific heat = 2506.2 joules perkgC
or can be written as 2.506 joules per gC
Hope this answer helps please eave a rating :)
Get Answers For Free
Most questions answered within 1 hours.