Show your work and include units with your answers for full credit. 150 g of col

Question

Show your work and include units with your answers for full credit. 150 g of cold lemonade (same specific heat as water) at 3.0 degree C is poured into a glass whose specific heat is 840 j / (kg C degree). The glass has just been washed in a dishwater, so its temperature is 75 degree C. After a few minutes, the lemonade and the glass reach an equilibrium temperature of 10 degree C. Assume an insulating container so that no heat is lost to the environment. (a) How much heat (in J) is gained by the lemonade? What is the mass of the glass (in g)?

Explanation / Answer

a)

Swater = 4.186 J /gm*K

heat gained = m*S*del T

heat gained = 150 * 4.186 * (10 - 3)

heat gained = 4395.3 J

the heat gained by the lemonade is 4395.3 J

b)

heat lost by glass = heat gained by the glass

heat lost by glass = 4395.3 J

heat lost by glass is 4395.3 J

c)

Now, as heat lost by glass = m * Sglass * del T

4395.3 = m*0.840 * (75 -10 )

m = 80.5 gm

the mass of glass is 80.5 gm

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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