Question

Water is being boiled in the flat bottom kettle placed on a stove. The area of the bottom is 3000 cm² and the thickness is 2 mm. If the amount of heat produced is 1gm/min, Calculate the temperature difference between inner and outer surface of the water.(k=0.5 cal/°C m s, L=540 cal/gm)​

Answer 1

Answer

Let the temperature difference be t,

Let the temperature difference be t, so the rate of heat conduction is given as Q/t=KAt/l where K=0.5cal is the thermal conductivity, A=300cm

Let the temperature difference be t, so the rate of heat conduction is given as Q/t=KAt/l where K=0.5cal is the thermal conductivity, A=300cm 2

Let the temperature difference be t, so the rate of heat conduction is given as Q/t=KAt/l where K=0.5cal is the thermal conductivity, A=300cm 2 is the area of surface, l=0.2cm is the thickness.

Let the temperature difference be t, so the rate of heat conduction is given as Q/t=KAt/l where K=0.5cal is the thermal conductivity, A=300cm 2 is the area of surface, l=0.2cm is the thickness.Now rate of vaporization is m

Let the temperature difference be t, so the rate of heat conduction is given as Q/t=KAt/l where K=0.5cal is the thermal conductivity, A=300cm 2 is the area of surface, l=0.2cm is the thickness.Now rate of vaporization is m l

Let the temperature difference be t, so the rate of heat conduction is given as Q/t=KAt/l where K=0.5cal is the thermal conductivity, A=300cm 2 is the area of surface, l=0.2cm is the thickness.Now rate of vaporization is m l

Let the temperature difference be t, so the rate of heat conduction is given as Q/t=KAt/l where K=0.5cal is the thermal conductivity, A=300cm 2 is the area of surface, l=0.2cm is the thickness.Now rate of vaporization is m l =1gm min

Let the temperature difference be t, so the rate of heat conduction is given as Q/t=KAt/l where K=0.5cal is the thermal conductivity, A=300cm 2 is the area of surface, l=0.2cm is the thickness.Now rate of vaporization is m l =1gm min −1

Let the temperature difference be t, so the rate of heat conduction is given as Q/t=KAt/l where K=0.5cal is the thermal conductivity, A=300cm 2 is the area of surface, l=0.2cm is the thickness.Now rate of vaporization is m l =1gm min −1 , so heat absorption rate is 1×540/60=9cals

Let the temperature difference be t, so the rate of heat conduction is given as Q/t=KAt/l where K=0.5cal is the thermal conductivity, A=300cm 2 is the area of surface, l=0.2cm is the thickness.Now rate of vaporization is m l =1gm min −1 , so heat absorption rate is 1×540/60=9cals −1

Let the temperature difference be t, so the rate of heat conduction is given as Q/t=KAt/l where K=0.5cal is the thermal conductivity, A=300cm 2 is the area of surface, l=0.2cm is the thickness.Now rate of vaporization is m l =1gm min −1 , so heat absorption rate is 1×540/60=9cals −1

Let the temperature difference be t, so the rate of heat conduction is given as Q/t=KAt/l where K=0.5cal is the thermal conductivity, A=300cm 2 is the area of surface, l=0.2cm is the thickness.Now rate of vaporization is m l =1gm min −1 , so heat absorption rate is 1×540/60=9cals −1 Equating the heat rate, we get 0.5×300×t/0.2=9⇒t=0.012

Let the temperature difference be t, so the rate of heat conduction is given as Q/t=KAt/l where K=0.5cal is the thermal conductivity, A=300cm 2 is the area of surface, l=0.2cm is the thickness.Now rate of vaporization is m l =1gm min −1 , so heat absorption rate is 1×540/60=9cals −1 Equating the heat rate, we get 0.5×300×t/0.2=9⇒t=0.012 o

Let the temperature difference be t, so the rate of heat conduction is given as Q/t=KAt/l where K=0.5cal is the thermal conductivity, A=300cm 2 is the area of surface, l=0.2cm is the thickness.Now rate of vaporization is m l =1gm min −1 , so heat absorption rate is 1×540/60=9cals −1 Equating the heat rate, we get 0.5×300×t/0.2=9⇒t=0.012 o C