Thermodynamic heat power problem

AI Thread Summary
To determine the power required for a carpet steamer to convert 2 kg of snow or ice at -10°C to steam in 5 minutes, the relevant thermodynamic equations include Q=mCΔT, Q=mHv, and Q=mHf. The calculations involve accounting for the energy needed to raise the temperature of the ice, melt it, and then convert the resulting water to steam. An initial attempt at solving the problem involved incorrect division of the given values, leading to an inaccurate result. The correct power needed for the steamer is calculated to be 20156 W. Understanding the thermodynamic principles and correctly applying the equations is crucial for solving such problems.
pasiman94
Messages
1
Reaction score
0

Homework Statement



Carpet steamer marketed for Greenland. Add up to 2 kg of snow or ice (no colder than -10°C) that will be converted to steam within 5 minutes. How powerful would the steamer need to be to make this claim? Know that 2108 J/kg K; Hf= 3.24*10^5 J/kg; Cwater= 4184 J/kg K; Hv = 2.26*10^6 J/kg.

Homework Equations



Q=mCΔT
Q=mHv
Q=mHf

The Attempt at a Solution


attempted to solve the equation by dividing all the givens by two and then adding them all up, however I know that it is not correct and the answer is P = 20156 W

Thanks
 
Physics news on Phys.org
pasiman94 said:

Homework Statement



Carpet steamer marketed for Greenland. Add up to 2 kg of snow or ice (no colder than -10°C) that will be converted to steam within 5 minutes. How powerful would the steamer need to be to make this claim? Know that 2108 J/kg K; Hf= 3.24*10^5 J/kg; Cwater= 4184 J/kg K; Hv = 2.26*10^6 J/kg.

Homework Equations



Q=mCΔT
Q=mHv
Q=mHf

The Attempt at a Solution


attempted to solve the equation by dividing all the givens by two and then adding them all up, however I know that it is not correct and the answer is P = 20156 W

Thanks

What was the logic behind your approach? Why do you think it failed?
 
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