How much time does it take to warm 2 litres water

  • Thread starter Thread starter mahima
  • Start date Start date
  • Tags Tags
    Time Water
AI Thread Summary
To determine the time required to heat 2 liters of water by 1°C with a 200V supply, one must first calculate the power output using the formula P = V²/R, where V is voltage and R is resistance. With a resistance of 5 ohms, the power at 200V is 8000 watts. Since it takes 10 minutes to heat 1 liter by 1°C at 100V, the energy required for 2 liters is double, meaning it would take 20 minutes at the original voltage. However, with the increased power at 200V, the time to heat 2 liters by 1°C would be significantly reduced. The final calculation shows that it would take 2.5 minutes to heat 2 liters of water by 1°C at 200V.
mahima
Messages
3
Reaction score
0
A resistance of 5(ohms) is connected to a 100V supply. If this resistance increase with temperature of 1 litre of water in 10 minutes by 1(oC-degree celsius), how much time it will take to increase temperature of 2 litres of water by 1(oC) if voltage is increased to 200V...

Can I have an answer with solution...
:confused:
 
Physics news on Phys.org
Thread moved to Homework Help Intro Physics.

mahaima, welcome to the PF. You must post homework/coursework questions in the Homework Help forum area of the PF, and you must show us your own work so far, before we can offer tutorial help.

How would you start this problem?...
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top