Liquid in a tin can

Classical Mechanics Level 4

v g v\text{ g} of liquid at 10 0 ∘ C 100 ^\circ C is added into a tin can at 0 ∘ C 0 ^\circ C weighing m g m\text{ g} . Given that the heat capacity of the liquid is c L J/K c_L\text{ J/K} and that of the tin can is c T J/K c_T\text{ J/K} , what is the equilibrium temperature θ \theta of the system?

Assume the liquid remains as liquid and the tin can remains solid.

θ = 100 c L c L + c T \theta=\frac{100c_L}{c_L+c_T} θ = 100 m c L m c L + v c T \theta=\frac{100mc_L}{mc_L+vc_T} θ = 100 v c L v c L + m c T \theta=\frac{100vc_L}{vc_L+mc_T} θ = 373 v c L v c L + m c T \theta=\frac{373vc_L}{vc_L+mc_T}

Wee Xian Bin by Wee Xian Bin , 25, Singapore

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1 solution

Wee Xian Bin
Jul 2, 2016

At thermal equilibrium, Q = c L ( 100 − θ ) = c T ( θ − 0 ) Q=c_L(100-θ)=c_T(θ-0) . Note that “heat capacity” has already taken into account the object’s mass (in contrast to specific heat capacity).

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