Answer:
Explanation:
There are a couple of ways you could do this.
The easiest is to use E*R1/(R1 + R2)
- E = 10 volts
- R1 = 590 ohms
- R2 = 840 ohms
So the result would be
E_590 = 10 * 590/(590 + 840)
E_590 = 10 * 590/ (1430)
E_590 = 4.13 volts rounded.
You could do this a slightly longer way.
R = 1430 (total ohms in series.
E = 10 volts
I = ???
I = E/R
I = 10 / 1430
I = 0.00699
Now use this current to figure out the voltage drop.
E = I * R
I = 0.00699 amps
R = 590 ohms
E = 0.00699 * 590
E = 4.13 volts
Pick the way of doing it you like best.
Answer:
Explanation:
The amount of heat released from water is equal to the sum of latent and sensible heats. Let suppose that water is initially at a temperature of 
. Then:
Finally, the amount of heat released from water is now computed by replacing variables:
![\Delta H_{tot} = (1\,mol)\cdot \left[\left(75.3\,\frac{kJ}{mol\cdot K} \right)\cdot (25^{\circ}C-0^{\circ}C)+ 6.025\,\frac{kJ}{mol} + \left(37.7\,\frac{kJ}{mol\cdot K} \right)\cdot (0 + 10^{\circ}C)\right]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Btot%7D%20%3D%20%281%5C%2Cmol%29%5Ccdot%20%5Cleft%5B%5Cleft%2875.3%5C%2C%5Cfrac%7BkJ%7D%7Bmol%5Ccdot%20K%7D%20%5Cright%29%5Ccdot%20%2825%5E%7B%5Ccirc%7DC-0%5E%7B%5Ccirc%7DC%29%2B%206.025%5C%2C%5Cfrac%7BkJ%7D%7Bmol%7D%20%2B%20%5Cleft%2837.7%5C%2C%5Cfrac%7BkJ%7D%7Bmol%5Ccdot%20K%7D%20%5Cright%29%5Ccdot%20%280%20%2B%2010%5E%7B%5Ccirc%7DC%29%5Cright%5D)
