**1. In an electrically heated home, the temperature of the ground in contact with a concrete basement wall is 13.2 **^{C}. The temperature at the inside surface of the wall is 21.7 ^{C}. The wall is 0.18 m thick and has an area of 8.0 m^{2}. Assume that one-kilowatt hour of electrical energy costs $0.10. How many hours are required for one dollar’s worth of energy to be conducted through the wall?

^{C}. The temperature at the inside surface of the wall is 21.7

^{C}. The wall is 0.18 m thick and has an area of 8.0 m

^{2}. Assume that one-kilowatt hour of electrical energy costs $0.10. How many hours are required for one dollar’s worth of energy to be conducted through the wall?

**Ans:**

Amount of heat produced per second, Q/t=kAdeltaT/L;

Where k is the thermal conductivity of the material, here=k=1.1 W/m^{O}c,

Q/t = (1.1*8*(21.7-13.2))/0.18

= 415.55 W.

The time required to produce energy of 1kWh=1kWh/415.55 W=3600*10^{3}J/(415.55J/s)

=8.6632*10^{3 }s.

Time taken for one dollar worth energy=8663.2/0.01=86632s=24.06 hrs.

**2. In an aluminum pot, 0.453 kg of water at 100 °C boils away in four minutes. The bottom of the pot is 3.86 × 10**^{-3}m thick and has a surface area of 0.0235 m^{2}. To prevent the water from boiling too rapidly, a stainless steel plate has been placed between the pot and the heating element. The plate is 1.90 × 10^{-3} m thick, and its area matches that of the pot. Assuming that heat is conducted into the water only through the bottom of the pot, find the temperature in degrees Celsius at the steel surface in contact with the heating element.

^{-3}m thick and has a surface area of 0.0235 m

^{2}. To prevent the water from boiling too rapidly, a stainless steel plate has been placed between the pot and the heating element. The plate is 1.90 × 10

^{-3}m thick, and its area matches that of the pot. Assuming that heat is conducted into the water only through the bottom of the pot, find the temperature in degrees Celsius at the steel surface in contact with the heating element.

**Ans: **

Change in temperature=delta T=QL/kAT, where Q=mL_{v} (L_{v}= latent heat of vaporization).

Delta T= mL_{v}L/kAT

=0.453kg*22.6*10^{5}J/kg*3.86*10^{-3} m/(240 W/mK*0.0235 m^{2}* 240 s)

=17.51°C

T_{aluminium} = 100 °C +17.51°C

=117.51°C

Similarly, for deltaT _{steel}= 24.63°C

**T _{steel}=142.14°C**

**3. A person eats a dessert that contains 180 Calories. (This “Calorie” unit, with a capital C, is the one used by nutritionists; 1 Calorie = 4186 J.) The skin temperature of the person is 36 °C and that of her environment is 18 °C. The emissivity of her skin is 0.84 and its surface area is 1.5 m**^{2}. How many hours would it take her to emit a *net radiant* energy from her body that is equal to the energy contained in this dessert?

^{2}. How many hours would it take her to emit a

*net radiant*energy from her body that is equal to the energy contained in this dessert?

**Ans: **

Net thermal energy radiated (as per Stephans Boltzmann law,)

Q _{tot}=eσA(T^{4}_{body}-T^{4}_{env})t…………………………………………………………….(1)

Given, Q _{tot}=180*4186 J

= 753480J…………………………………………………………….(2)

From equations (1) and (2) we can calculate t

t = Q _{tot/} eσA(T^{4}_{body}-T^{4}_{env})

**t=1.50 hrs.**