A well-insulated, rigid vessel contains 3 kg of saturated liquid water at 40°C. The vessel also contains an electrical resistor that draws 10 A when 50 V are applied. Determine the final temperature in the vessel after the resistor has been operating for 30 min. (Solve this problem using appropriate software.)

The final temperature in the vessel after the resistor has been operating for 30 min is 111.67°C

Explanation:

given information:

mass, m = 3 kg

initial temperature,  T₁ = 40°C

current, I = 10 A

voltage, V = 50 V

time, t = 30 min = 1800 s

Heat for the system because of the resistance is

Q = V I t

where

V = voltage (V)

I = current (A)

t = time (s)

Q = heat transfer to the system (J)

so,

Q = V x I x t

   = 50 x 10 x 1800

   = 900000

   = 9 x 10⁵ J

the heat transfer in the closed system is

Q = ΔU + W

where

U = internal energy

W = work done by the system

thus,

Q = ΔU + W

9 x 10⁵ = ΔU + 0, W = 0 because the tank is a well-insulated and rigid.

ΔU = 9 x 10⁵ J = 900 kJ

then, the energy change in the system is

ΔU = m c ΔT

ΔT = ΔU / m c, c = 4.186 J/g°C

     = 900 / (3 x 4.186)

     = 71.67°C

so,the final temperature (T₂)

ΔT = T₂ - T₁

T₂ = ΔT + T₁

    = 71.67°C + 40°C

    = 111.67°C

b. check your rearview mirror

explanation: