A 7.9-kg cube of copper (c Cu = 386 J/kg-K) has a temperature of 750 K. It is dr

Question

A 7.9-kg cube of copper (cCu = 386 J/kg-K) has a temperature of 750 K. It is dropped into a bucket containing 5.8 kg of water (cwater = 4186 J/kg-K) with an initial temperature of 293 K.

A 7.9-kg cube of copper (ccu 386 J/kg-K) has a temperature of 750 K. It is dropped into a bucket containing 5.8 kg of water Water 4186 J/kg-K) with an initial temperature of 293 K 1) What is the final temperature of the water-and-cube system? K Submit You currently have 2 submissions for this question. Only 10 submission are allowed. You can make 8 more submissions for this question. 2) If the temperature of the copper was instead 1350 K, it would cause the water to boil. How much liquid water (latent heat of vaporization 2.26 x 106 J/kg) will be left after the water stops boiling kg Submit You currently have 5 submissions for this question. Only 10 submission are allowed. You can make 5 more submissions for this question. 3) Let's try this again, but this time add just the minimum amount of water needed to lower the temperature of the copper to 373 K. In other words, we start with the cube of copper at 750 K and we only add enough water at 293 K so that it completely evaporates by the time the copper reaches 373 K. Assume the resulting water vapor remaining at 373 K. How much water do we need? kg Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question.

Explanation / Answer

(1)let the final temperature =T

Heat lost by copper= heat gained by water

m1 c1 (750-T) = m1 c2(T-293)

7.9(386)(750-T)=5.8(4186)(T-293)

T= 344 K.

(2) let m= mass of water that changes into vapours

Heat lost by copper= heat used for boling the 5.8 kg of water + heat used for vapourization of m kg

m1 c1 (1350-373) = m1 c2(373-293) + m L

L= latent heat of vapourization

7.9(386) (1350-373) = 5.8 (4186)(373-293) + m (2.26x10^6)

m=0.46 kg

so mass of liquid water left= 5.8-0.46

mass of liquid water left=5.34 kg.

(3)heat given out by copper= heat gained by water + heat required to change that water into vapor

7.9(386) (750-373) = m (4186)(373-293) + m (2.26x10^6)

1149624=2594880 m

m= 0.443 kg.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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