Respuesta :
Explanation:
It is given that,
Mass of boiled water, m = 4.5 kg
Radius of the pot bottom, r = 6.5 cm = 0.065 m
Thickness of the pot, t = 2 mm = 0.002 m
We need to find the temperature of the heating element in contact with the pot.
Heat transferred is from pot to water is given by :
[tex]H=\dfrac{kA(T_2-T_1)}{l}[/tex]
[tex]\dfrac{Q}{t}=\dfrac{kA(T_2-T_1)}{l}[/tex]
Tā is the boiling temperature of water, Tā = 100ā° C
[tex]T_2=\dfrac{Q}{kAt}+T_1[/tex], q = m L, L is the latent heat of vaporization
[tex]T_2=\dfrac{mL}{kAt}+T_1[/tex],
k is the thermal conductivity of water, [tex]k=390\ Jm^{-1}s^{-1}(^oC)^{-1}[/tex]
[tex]L=2.26\times 10^6\ Ā Jkg^{-1}[/tex]
[tex]T_2=\dfrac{4.5\times 2.26\times 10^6}{390\times \pi\times (0.065)^2\times 120}+100[/tex]
[tex]T_2=1737.18\ ^oC[/tex]
Hence, this is the required solution.
