Question

A sample of water with a mass of 726.2 g undergoes a temperature change from 58.32 ˚C to 40.50 ˚C. What is the heat change in Joules? (Specific Heat of H2O (l) = 4.184 J/g∙˚C)

Answer 1

To calculate the heat change (Q) in Joules, we can use the formula:

Q = m * c * ΔT

Where:
- Q is the heat change in Joules
- m is the mass of the water (in grams)
- c is the specific heat capacity of water (4.184 J/g∙˚C)
- ΔT is the change in temperature (final temperature - initial temperature)

Given:
- Mass (m) = 726.2 g
- Specific heat (c) = 4.184 J/g∙˚C
- Initial temperature = 58.32 ˚C
- Final temperature = 40.50 ˚C

First, calculate the change in temperature (ΔT):
ΔT = Final temperature - Initial temperature
ΔT = 40.50 ˚C - 58.32 ˚C
ΔT = -17.82 ˚C

Now, plug the values into the formula:
Q = (726.2 g) * (4.184 J/g∙˚C) * (-17.82 ˚C)

Calculate:
Q ≈ (726.2) * (4.184) * (-17.82)
Q ≈ -53,949.15 J

The heat change is approximately -53,949.15 Joules. Since the temperature decreased, the heat change is negative, indicating heat loss.