Calculating Energy Released in Water Temperature Change

AI Thread Summary
To calculate the energy released as water changes from 20°C to -10°C, the equation used is Energy = m_w c_w Δθ_w + m_w l_f + m_i c_i Δθ_i, where m_w is the mass of water, c_w is the specific heat of water, l_f is the latent heat of fusion, m_i is the mass of ice, and c_i is the specific heat of ice. The temperature change for water is from 20°C to 0°C, and for ice, it is from 0°C to -10°C. The discussion emphasizes the importance of correctly identifying the temperature changes and phases involved in the calculation. Participants express confusion about setting up the equation and plugging in the correct values. Understanding the phases and temperature changes is crucial for solving the problem accurately.
tornzaer
Messages
77
Reaction score
0

Homework Statement


What is the energy released as water changes temperature from 20C to -10C?


Homework Equations


mct+lfm+mct


The Attempt at a Solution


My physics exam is day after tomorrow, however, I have no clue about this question. I was away for a week during these lessons. Can someone please walk me through the question so I can understand.

Thank you very much.
 
Physics news on Phys.org
Assuming no heat losses. All the energy released goes into reducing the temperature from 20C to 0C and changing the water's phase and then lowering the temperature of the ice from 0C to -10C
 
Well yes, but how would I set up the equation?
 
Energy=m_w c_w \Delta \theta_w + m_w l_f +m_ic_i \Delta \theta_i

i=ice,w=water.
 
Can you please plug in the digits? I am confused about the temperatures.
 
rock.freak667 said:
Energy=m_w c_w \Delta \theta_w + m_w l_f +m_ic_i \Delta \theta_i

i=ice,w=water.

for the change in temperature of the water it is (20-0)C and for ice it is (0-(-10))=10C.
 

Thread 'Errors and significant figures'

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...
View full post »
Thread 'A cylinder connected to a hanging mass'

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...
View full post »

Similar threads

Why Is Energy Released in Nuclear Decay Equal to Change in Binding Energies?
Replies
13
Views
1K
Change in Entropy When Mixing Water at Different Temperatures
Replies
5
Views
1K
What is the temperature change of a bullet upon impact?
Replies
28
Views
3K
Energy of a water tank with compressed air
Replies
5
Views
1K
Incorrect Calculation of Time for Reactor Energy Phase Change
Replies
1
Views
2K
Effect of temperature on surface tension
Replies
2
Views
1K
Temperature change of Water after dropping a ball into it
Replies
10
Views
3K
How Does a Dropped Ball Affect the Temperature of Water?
Replies
6
Views
3K
Calculate internal temperature and energy changes
Replies
1
Views
884
Correlating albedo with temperature?
Replies
6
Views
3K

Hot Threads

Recent Insights

Back
Top