To start a great night of doing physics homework, you sit down to pour yourself

Question

To start a great night of doing physics homework, you sit down to pour yourself a good cup of coffee. Your coffee mug has a mass 121.0 g and a specific heat of 1089.0 J/kg K. The mug starts out at room temperature (24.0 o C). Your coffee has an initial temperature of 81.7 oC and has the same specific heat as water (4186.0 J/kg K). 2) What is the final temperature of coffee and mug once they come to thermal equilibrium? Tfinal = oC 3) Now lets assume that instead of the 121.0 g of coffee, you pour in 240.0 g of coffee. What is the final temperature of the coffee and mug? (Again, assume that you loose no heat to the outside.) Tfinal = oC 4) Now lets say that along with the 240.0 g of coffee, you pour in 11.7 g of cream in your mug. The cream has an initial temperature of 4.7 oC and also has the same specific heat as water. What is the final temperature of the coffee, cream and mug? (Again, assume that you loose no heat to the outside.)

Explanation / Answer

Q=mcT

The question did not specify the mass of the coffee initially, but it made reference in the subsequent part to be 121 g as well , so i'm just gonna use that value

Heat absorbed by mug = Heat given out by coffee

(0.121)*(1089)*(T-24)= 4186*(0.121)(81.7 -T)

solving, T=69.8 C

Do the same thing, but use mass of coffe = 0.24kg, T now become 75.0 C

Coffee with cream:

(0.121)*(1089)*(T-24)+ (0.117)*(1089)*(T-4.7)=4186*(0.24)(81.7 -T)

solving T= 67.9 C

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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