8.0 g of cream at 17.0 °C are added to an insulated cup containing 150.0 g of coffee at 73.2 °C. Calculate the equilibrium temperature of the coffee. You may assume no heat is lost to the cup or surroundings, and that any physical properties of cream and coffee you need are the same as those of water. Be sure your answer has the correct number of significant digits

The equilibrium temperature of the coffee is 70.4  °C

Explanation:

Step 1: Data given

Mass of cream = 8.0 grams

Temperature of the cream = 17.0°C

Mass of the coffee = 150.0 grams

Temperature of the coffee = 73.2 °C

C = respective specific heat of the substances( same as water) = 4.184 J/g°C

Step 2: Calculate the equilibrium temperature

m(cream)*C*(ΔT) = -m(coffee)*c*(T2-T1)

⇒with m(cream) = the mass of cream = 8.0 grams

⇒with C = the specific heat of cream = 4.184 J/g°C

⇒with ΔT(cream) = T2 - T1 = T2 - 17.0°C

⇒with m(coffee) = the mass of coffee = 150.0 grams

⇒with C(coffee) = the specific heat of coffee = 4.184 J/g°C

⇒with ΔT(coffee) = T2 - T1 = T2 - 73.2 °C

8.0 g* 4.184 J/g°C *(T2 - 17.0°C) = -150.0g *4.184 J/g°C*(T2-73.2°C)

8.0 g*(T2 - 17.0°C) = -150.0g *(T2-73.2°C)

8T2 - 136 = -150T2 +10980

158T2 = 11116

T2 = 70.4 °C

The equilibrium temperature of the coffee is 70.4  °C

false

because it does not have the potential to be entirely used up