Thermal+capacity+Question+A

Given information; Total thermal capacity of calorimeter, heater, stirrer and thermometer: C= 107 J K -1 Mass of liquid: m=0.241kg Potential difference: V=12.2V Current: I=3.40A Initial rate of rise of temperature: ΔΘ/Δt =3.7 x 10 -2Ks -1

(a) Power = VI = 12.2 x 3.4 = 41.48 Js -1

(b) Assume there is no heat loss to the surroundings. From equations; Q= Pt, Q = C(ΔΘ) + mc(ΔΘ)

Total P(Δt) = C(ΔΘ) + mc(ΔΘ)

Therefore, P =C(ΔΘ/Δt) +mc(ΔΘ/Δt) ; 41.48 = 107 (3.7 x 10 -2 ) + 0.241 c (3.7 x 10 -2 )

Thus, c= 4.21 x 10 3 J kg -1 K