Suppose v = 24 v, i = 0.1 a, mw = 46 gm, cw = 4.18 j/g∘k, mr = 8 g, and cr = 3.7 j/g∘k. if the water is initially at room temperature, how long will it take for the water to heat up 2 k? (hint: dt/dt is approximately equal to δt/δt .)

16 seconds

Explanation:

Solution.

We are to determine the time required to heat a mass of water with a temperature change of 2 K, with an electric resistance of 24 V and current intensity of 1A.

Assumptions:

Steady state condition, since the temperature change is small.

All of the electric energy is converted into heat by the resistance.

Analysis.

When the steady state conditions are reached, the heat loss of the resistance will be equal to the heat generation rate of the electric resistance; meaning,

Q’=Q/ Δt =E’,generated=V*i

Where

Q is the heat transferred to the water, evident as latent heat. Then,

Q=m,w*Cp,w*ΔT/Δt=V*i

Then,

Δt= m,w*Cp,w*ΔT/(V*i)

Replacing values

Δt=(46 g)*(4.18 J/g K)*(2 K)/(24 J/A s *1A)

Δt=16 s

Discussion.

For this problem, we could also consider a transient state condition, that includes the temperature profile change within the resistance (given that we have the mass and specific heat information). However, we would require more information about the geometry in order to evaluate the heat transfer coefficients to the water. Also, the heating time of 16 seconds is quite reasonable given the mass of water and the electric resistance of 24 V.

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6 times more

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