Heat and Internal Energy and Heat and Temperature Change: Specific Heat Capacity

Axoren
Messages
17
Reaction score
0

Homework Statement


When you take a bath, how many kilograms of hot water (49°C) must you mix with cold water (12°C) so that the temperature of the bath is 36°C? The total mass of water (hot plus cold) is 191 kg. Ignore any heat flow between the water and its external surroundings.

Temperature of the Cold Water: 12 C
Temperature of the Hot Water: 49 C
Temperature of the Total Water: 36 C
Mass of the Cold Water: y, 191-x
Mass of the Hotwater: x, 191-y
Mass of the Total Water: 191 kg

Homework Equations


Q =mcΔT
Where Q is Joules of Energy, m is Mass of the substance (kg), c is the specific heat coefficient, ΔT is change in Temperature (in Celsius)

The Attempt at a Solution


Total Q = 191 kg * 4186 J/kgC * 36 C
Total Q = 28782936 J
Hot Q = X kg * 4186 J/kgC * 13 C
Cold Q = (191 - X) kg * 4186 J/kgC * 24 C
28782936 J = Hot Q + Cold Q
28782936 J = (X kg * 4186 J/kgC * 13 C) + ((191 - X) kg * 4186 J/kgC * 24 C)
28782936 J = (X kg * 54418 J/kg) + ((191 - X) kg * 100464 J/kg)
28782936 J = 54418X J + 19188624 J - 100464X J
28782936 J = 19188624 J - 46046X J
9594312 J = 46046X J
208.36 kg = X (Answer only needs to be to 2 decimal places)

^This does not make sense, because the mass of the hot water is larger than the mass of the total water. Assistance?
 
Last edited:
Physics news on Phys.org
Axoren said:

The Attempt at a Solution


Total Q = 191 kg * 4186 J/kgC * 36 C
Total Q = 28782936 J
Hot Q = X kg * 4186 J/kgC * 13 C
Cold Q = (191 - X) kg * 4186 J/kgC * 24 C
28782936 J = Hot Q + Cold Q
28782936 J = (X kg * 4186 J/kgC * 13 C) + ((191 - X) kg * 4186 J/kgC * 24 C)
28782936 J = (X kg * 54418 J/kg) + ((191 - X) kg * 100464 J/kg)
28782936 J = 54418X J + 19188624 J - 100464X J
28782936 J = 19188624 J - 46046X J
9594312 J = 46046X J
208.36 kg = X (Answer only needs to be to 2 decimal places)

^This does not make sense, because the mass of the hot water is larger than the mass of the total water. Assistance?
Total Q = 0 .

So the equation should be Q_h + Q_c=0

AM
 
Thank you very much! I don't know how I missed that.
 
Hey I really need this answered I've been working on it for 3 hours help.
Question: You want to take a bath with the water temperature at 35.0 degrees C. The water temperature is 38 degrees C from the hot water tap and 11 degrees C from the cold water tap. You fill the tub with a total of 187 kg of water. How many kilograms of water from the hot water tap do you use?
Please help.
 
Just kidding I'm stupid. i figured it out... sorry for wasting your time.
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top