Energy to raise temperature of water

AI Thread Summary
To determine how much freshly boiled water is needed to raise the temperature of 0.500 kg of water from 40.0°C to 65.0°C, the calculations show that approximately 0.357 kg of water at 100°C is required. However, there is a critical oversight regarding the capacity of the pot, as adding this amount would cause it to overflow. The discussion highlights the importance of maintaining accurate units throughout the calculations, emphasizing that proper tracking of units is essential for clarity. The final conclusion stresses that the water must be added to a larger pot to avoid overflow, indicating a practical consideration in the problem-solving process. Accurate calculations and practical application are both crucial in this scenario.
Dazed
Messages
26
Reaction score
0
Just looking for someone to check my work on this question before I submit it to the teacher. Any help is appreciated. :)

3. A 0.500 kg pot of hot water for tea has cooled to 40.0 OC. How much freshly boiled water must be added (at 100 OC) to raise the temperature of the tea water to a respectable 65.0 OC?

Let n represent the unknown mass of water to be added.

Starting water + Added water = total water

40c x 0.5 kg + 100c x n = 65 (0.5kg + n)

20 kg + 100 n = (32.5kg + 65n)

52.5 kg = 165 n

1 kg = 3.1n

n = 1/3.1

n = 0.32 kg

To raise the temperature of the tea to 65 degrees, 0.32kg of 100c water must be added.


~Dazed
 
Physics news on Phys.org
No, you messed up the third step. Subtract 65n from both sides, and 20 kg, to get
35n = 12.5kg
You can finish it from there.

A small additional note: your units aren't complete, when you multiply temperature times mass, the units are c-kg. Won't matter in the end, but keeping accurate track is a good habit.
 
Thank you for your reply. :)

I think I've correctly fixed this as per your suggestions, I used the units c/kg instead of c-kg, I assumed that was a typo. If c-kg is correct please tell me and I'll change it. Also please tell me if the new answer is correct or of I messed up somewhere again. :)

3. A 0.500 kg pot of hot water for tea has cooled to 40.0 OC. How much freshly boiled water must be added (at 100 OC) to raise the temperature of the tea water to a respectable 65.0 OC?

Let n represent the unknown mass of water to be added.

Starting water + Added water = Total water

(40c x 0.5 kg) + (100c * n) = 65c(0.5kg + n)

20c/kg + 100n = (32.5c/kg + 65n)

Subtract 65n from both sides
20c/kg + (100n - 65n) = 32.5c/kg + (65n – 65n)

20c/kg + 35n = 32.5c/kg

Subtract 20kg from both sides
(20kg – 20kg) + 35n = (32.5kg – 20kg)

35n = 12.5kg

n = 12.5kg / 35

n = 0.357kg

To raise the temperature of the tea to 65 degrees, 0.357kg of 100c water must be added.
 
Correct answer. You're units are still a little confused. If you multiply 40c x 0.5 kg, you get 20 c kg (whatever that is!). Then when you get down to
35c * n = 12.5 c kg
and solve, you get
n =12.5 c kg / 35 c
and the answer has units of kg as you knew it must.
 
NO. If you added 0.357kg of water to the pot it would overflow. You would have to add both parts of the water to a 0.900 kg pot and then return 0.500 kg to the original pot.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top