A device similar to the one diagrammed below has been demonstrated in class. A f

Question

A device similar to the one diagrammed below has been demonstrated in class. A fan motor draws energy from the heat flow between two temperature reservoirs. In the situation below two cups hold ice water and hot water respectively. In order to analyze the device consider a system (for energy tracking purposes) that consists of only the hot water and the ice water (both the ice and the water). Assume that no heat flows out of or into this system, but that energy leaves the system in the form of work done to turn the fan. ( We are ignoring any energy absorbed by the metal parts etc.). As a result Delta Eint = W where W is the work being done, which in this case is negative as it represents energy leaving the system into the fan. During one experiment using the device, 100 grams of hot water, initially at 80degree C, are used along with 100 grams of 0 degree C liquid water mixed with 50 grams of ice. At the end of a time interval where -1800 J of work are done to turn the fan, the cold reservoir is still at 0 degree C but it contains 122 grams of liquid water and 28 grams of ice. Using the equation Delta Eint = W for the system as defined determine the temperature of the hot water at the end of the time interval. The fan is removed from the reservoirs at the end of this period isolating the cups from each other. If the fan was disconnected (or electrically shorted) but a heat flow still occurred through the device between the hot and cold reservoirs, would the amount of ice melted be more or less than with the fan on? Explain your reasoning.

Explanation / Answer

(a) For the (liquid water+ice) system,

Initial mass of 0'C water = 100g

Initial mass of 0'C ice = 50g

At the end of the time interval, mass of ice left = 28g

So, mass of ice melted into water = 50-28 = 22g

Thus, heat absorbed by the (liquid water+ice) system during the interval = heat absorbed to melt 22g ice into water = mass of ice melted*latent heat of fusion of ice = 22g * 333 (J/g) = 7326 Joules

Change in internal energy = heat absorbed by (liquid ice+water) system - heat released by hot water

Now, heat released by hot water = mass of hot water * specific heat of water * change in temp

= 100 * 4.18 * (80-T)

Here, T is the final temperature of the hot water

So, chage in internal energy = 7326 - 100 * 4.18 * (80-T)

Work done = -1800 J

Now, change in internal energy = work done

7326 - 100 * 4.18 * (80-T) = -1800

Solving, we get : T = 58.17 'C

(b) If the fan had been off, no work would have been done on the fan. Instead, the 1800 J energy which went to the fan would also have been given to the (liquid water+ice) system

So, more amount of ice would have melted, corresponding to the extra 1800 J it would have received which initially went into the fan.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.