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 137 g and a specific heat of 1089 J/kg K. The mug starts out at room temperature (23.8 o C). Your coffee has an initial temperature of 79.8 oC and has the same specific heat as water (4186 J/kg K).

1) Let's say you pour enough coffee into the mug so that the mass of coffee is equal to the mass of the mug. If we assume that no heat is lost to the outside room, once the mug and coffee reach the same temperature, will that temperature be closer to the initial temperature of the coffee or the initial temperature of the mug? (Notice that the mass of the coffee and the mass of the mug are the same.)
2) What is the final temperature of coffee and mug once they come to thermal equilibrium?
Tfinal = ____________ degrees c
3) Now lets assume that instead of the 137 g of coffee, you pour in 225 g of coffee. What is the final temperature of the coffee and mug? (Again, assume that you loose no heat to the outside.)
Tfinal = ____________ degrees c
4) Now lets say that along with the 225 g of coffee, you pour in 11.7 g of cream in your mug. The cream has an initial temperature of 5.2 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.)
Tfinal = _____________ degrees c

Answer 1

1) The temperature will be closer to water

2) T = 68.239°C

3) T = 72.142°C

4) T = 69.266 °C

Step-by-step explanation:

1)

The temperature will be closer to water because the heat capacity of water > heat capacity of coffee.

2)

137(1.089)(T - 23.8) = 137(4.186)(79.8 - T)

⇒(1.089)(T - 23.8) = (4.186)(79.8 - T)

⇒1.089 T - 25.9182 = 334.0428 - 4.186 T

⇒1.089 T + 4.186 T = 334.0428 + 25.9182

⇒5.275 T = 359.961

⇒ T = 68.239°C

3)

137(1.089)(T - 23.8) = 225(4.186)(79.8 - T)

⇒(149.193)(T - 23.8) = (941.85)(79.8 - T)

⇒149.193 T - 3550.7934 = 75159.63 - 941.85 T

⇒149.193 T + 941.85 T = 75159.63 + 3550.7934

⇒1091.043 T = 78710.4234

⇒ T = 72.142°C

4)

137(1.089)(T - 23.8) + 11.7(4.186)(T - 5.2)= 225(4.186)(79.8 - T)

⇒(149.193)(T - 23.8) + 48.9762(T - 5.2) = (941.85)(79.8 - T)

⇒149.193 T - 3550.7934 + 48.9762 T - 254.67624= 75159.63 - 941.85 T

⇒149.193 T + 941.85 T + 48.9762 T = 75159.63 + 3550.7934 + 254.67624

⇒1091.043 T + 48.9762 T = 78710.4234 + 254.67624

⇒1140.0192 T = 78965.09964

⇒ T = 69.266 °C