Answer
47 minutes
Work Step by Step
Given:
Volume: V = 1 gal=0.133 ft^3
Initial Temperature: T_1 = 68 °F
Specific Heat: c=0.94 Btu/(lb °R)
Density of Milk: ρ=64 lb/(ft^3 )
Rate of heat transfer: Q ̇=-0.08 (Btu/s) (negative sign indicates energy is leaving the system)
Find:
Time required for the milk to cool to 40°F (T_2 )
By the first law of thermodynamics in rate form
du/dt=Q ̇-W ̇ (1)
Since there are no moving parts,W = 0.
du/dt=Q ̇-W ̇ (2)
du/dt=Q ̇ (3)
Separate variable from equation (3) and integrate
du=Q ̇dt (4)
∫du=∫Q dt (5)
since Q ̇ is constant,
∆u= Q ̇∆t (6)
And ∆u= mc(T_2-T_1 )=ρVc(T_2-T_1 ) (7)
Solve for ∆t and plug in values
∆t= ∆u/Q ̇ =ρVc(T_2-T_1 )/Q ̇ =(64 lb/(ft^3 )*0.133 ft^3*0.94 Btu/(lb °R)(40℉-68℉))/(-0.08 Btu/s)
∆t=2800 s≈47 minutes
