Question

If 100.0 g of water changes temperature from 40.0 °C to 35.0 °C, at constant pressure, what is the change in enthalpy for this sample of water? (The specific heat capacity of water is 4.184 J/g °C). Assume no heat transferred to surroundings.

a. 0 -2.09 kJ
b. -9.37 kJ
c. 2.09 kJ
d. 9.37 kJ

Answer 1

The correct answer is a. -2.09 kJ.The change in enthalpy for the cooling of 100.0 g of water from 40.0 °C to 35.0 °C is calculated using the formula ΔH = m * c * ΔT. Given the specific heat capacity of water is 4.184 J/g °C, the enthalpy change is -2.09 kJ.

Step-by-step explanation:

To determine the change in enthalpy (ΔH) for the cooling of 100.0 g of water from 40.0 °C to 35.0 °C, we can use the formula ΔH = m * c * ΔT, where 'm' is the mass of the water, 'c' is the specific heat capacity of water, and ΔT is the change in temperature. Given that the specific heat capacity of water is 4.184 J/g °C, and the mass (m) is 100.0 g, we can calculate the change in enthalpy as follows:

ΔH = (100.0 g) * (4.184 J/g °C) * (35.0 °C - 40.0 °C)

ΔH = (100.0 g) * (4.184 J/g °C) * (-5.0 °C)

ΔH = -2092 J

ΔH = -2.092 kJ because 1 kJ = 1000 J

Therefore, the change in enthalpy for this sample of water is -2.09 kJ, and the correct answer is a. -2.09 kJ.