The answer is B. Below freezing
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
ΔG⁰ = ΔH⁰ - T ΔS⁰
ΔG⁰ : Standard free energy of formation of acetylene
ΔH⁰ : Standard enthalpy of formation (226.7 kJ/mol)
ΔS⁰ : Standard entropy change (58.8 J / K. mol)
T : Temperature 25°C = 298 K (room temperature)
ΔG⁰ = 226.7 - (298 x 58.8 x 10⁻³) = 209.2 kJ /mol
