Question

15.0 g of cream at 10.0 ℃ are added to an insulated cup containing 150.0 g of coffee at 78.6 °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 3 significant digits. _____________⁰C

Answer 1

Answer:

The equilibrium temperature of the coffee is 72.4 °C

Explanation:

Step 1: Data given

Mass of cream = 15.0 grams

Temperature of the cream = 10.0°C

Mass of the coffee = 150.0 grams

Temperature of the coffee = 78.6 °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*(T2-T1) = -m(coffee)*c*(T2-T1)

15.0 g* 4.184 J/g°C *(T2 - 10.0°C) = -150.0g *4.184 J/g°C*(T2-78.6°C)

62.76T2 - 627.6 = -627.6T2 + 49329.36

690.36T2 = 49956.96

T2 = 72.4 °C

The equilibrium temperature of the coffee is 72.4 °C