A 1.2 kg aluminum frying pan is taken from a stove and placed in a kitchen sink.

Question

A 1.2 kg aluminum frying pan is taken from a stove and placed in a kitchen sink. The faucet is dripping cold water (10°C) onto the pan, and the drops, of course, vaporize when they hit the hot pan.

    (A) if each drop is 3 grams, how much energy is used in vaporizing it?

    (B) How much does the pan temperature decrease with each drop?

    (C) If there is a drop every second, what is the rate of cooling (in watts) from the water?

    (D)   Does the water significantly improve the rate at which the pan would cool by radiation? Support your answer with a calculation.   You may take the emissivity of the pan as 0.3, and consider it as a disk of area 4 x 10-3 m2.  

Explanation / Answer

Solution: When a drop fall on the hot pan, its temperature increases from 10oC to 100oC and then it goes into to gaseous state from liquid state.

The specific heat of water c = 4180 J/kg.oC

Heat of vaporization for water L = 2256 kJ/kg = 2.256*106 J/kg

Mass of each drop is m = 3 grams = 3*10-3 kg

PART (A)

The heat needed to increase the temperature of water drop from Ti = 10oC to Tf = 100oC is given by,

Q1 = c*m*( Tf - Ti)

Q1 = (4180 J/kg.oC)*( 3*10-3 kg)*( 100oC - 10oC)

Q1 = 1128.6 J

The heat needed to vaporize the water drop at 100oC is given by,

Q2 = L*m

Q2 = (2.256*106 J/kg)*( 3*10-3 kg)

Q2 = 6768.0 J

Thus the total energy needed to vaporize the

Q = Q1 + Q2

Q = 1128.6 J + 6768 J

Q = 7896.6 J

----------------------------------------------------------------------------------------------------------------------------------------------

PART(B)

Specific heat of aluminum pan is c = 900 J/kg. oC

The mass of the aluminum pan is m = 1.2 kg

As the each drop falls on the drop and vaporizes, it takes 7896.6 Joules of heat from the heated pan, thus its temperature decreases by T

Q = c*m*T

T = Q/c*m

T = (7896.6 J)/((900 J/kg. oC)*(1.2kg)

T = 7.3116 oC

Thus with each drop fallen, the temperature of the pan decreases by 7.3 oC

------------------------------------------------------------------------------------------------------------------------------------------

PART(C)

Since each drop is falling each second, it withdraws 7896.6 Joules of heat each second. Thus the rate of cooling,

P = Q/t

P = (7896.6 J))/(1s)

P = 7896.6 J/s

P = 7896.6 Watts

------------------------------------------------------------------------------------------------------------------------------------------

PART(D) The water can vaporize only when the temperature of the pan is above 100 oC.

WE notice that, as long as the temperature of pan is above 100 oC, its heat loss is constant, that is   7896.6 Watts.

We can take two temperatures of the pan say 300 oC and compute the heat loss due to the radiation.

= 0.3 emissivity for the pan,

A = 4*10-4 m2, area of the pan,

= 5.6704*10-8 W/m2K4

Thus the rate at which pan cools at T = 300 oC = (300 +273)K = 573 K

Prad = **A*T4

Prad = (5.6704*10-8 W/m2K4)*(0.3)*(4*10-3 m2)*(573 K)4

Prad = 7.3352 Watt

Thus we note that the heat loss at 300oC is just 7.332 Watt as compared to the heat loss due to the vaporization of water (7896.6 watts)

Further, when the temperature drops, say 200oC; Prad also decreases as it is directly dependent on T4

Thus water significantly improves the rate at which pan would cool by radiation.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.