Answer:
The equilibrium temperature of the coffee is 70.4 °C
Step-by-step 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