Question

When 14.8 g KOH is dissolved in 85.6 g of water in a coffee-cup calorimeter, the temperature rises from 19.3 °C to 32.76 °C. What is the enthalpy change per gram of potassium hydroxide dissolved in the water?

Answer 1

Answer:

This question is incomplete, the complete question is; assuming that the solution has a specific heat of 4.18 J/g°C

The answer is 381.67 J/g

Explanation:

Enthalpy change denoted by ΔH can be calculated using the formula;

ΔH = m × c × ΔT

Where; m= mass of reactants

c= specific heat

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 32.76 - 19.3

ΔT = 13.46 °C

mass of reactants= 85.6 + 14.8 = 100.4g, c = 4.18J/g°C

Hence; ΔH = m × c × ΔT

ΔH = 100.4 × 4.18 × 13.46

ΔH = 5648.78J

Enthalpy change per gram of potassium hydroxide dissolved in the water is;

ΔH = 5648.78/14.8

ΔH = 381.67 J/g

Answer 2

The enthalpy change per gram of potassium hydroxide dissolved in the water is 382 J/g

How to determine the heat

• Mass (M) = 14.8 + 85.6 = 100.4 g
• Initial temperature (T₁) = 19.3 °C
• Final temperature (T₂) = 32.76 °C
• Change in temperature (ΔT) = 32.76  – 19.3 = 13.46 °C
• Specific heat capacity (C) = 4.184 J/gºC
• Heat (Q) =?

Q = MCΔT

Q = 100.4 × 4.184 × 13.46

Q = 5654.19 J

How to determine the enthalpy change ΔH

• Heat (Q) = 5654.19 J
• Mass of KOH (m) = 14.8 g

• Enthalpy change (ΔH) =?

ΔH = Q / g

ΔH = 5654.19 / 14.8

ΔH = 382 J/g

Learn more about heat transfer:

https://brainly.com/question/10286596