If 720.0 mL of liquid water absorbs 80.0 kJ of heat, what will be its increase in temperature?
Please refer to your next post, which I saw first.
Sra
To determine the increase in temperature of the water, we need to use the specific heat capacity of water to calculate the amount of energy required to raise the temperature.
The specific heat capacity of water is approximately 4.18 J/g°C (joules per gram per degree Celsius). However, since the given values are in milliliters (mL) and kilojoules (kJ), we need to convert them to grams and joules, respectively.
First, let's convert the volume of water from milliliters to grams. The density of water is approximately 1 g/mL, so the mass (m) of the water will be equal to its volume (V):
m = V = 720.0 mL
Next, let's convert the 80.0 kJ of heat to joules. Since 1 kJ = 1000 J, we can multiply the given value by 1000 to convert it:
Q = 80.0 kJ = 80,000 J
Now, we can use the formula:
Q = m * c * ΔT
Where:
- Q is the amount of heat energy absorbed (in joules),
- m is the mass of the substance (in grams),
- c is the specific heat capacity of the substance (in J/g°C),
- ΔT is the change in temperature (in °C).
Rearranging the formula to solve for ΔT:
ΔT = Q / (m * c)
Plugging in the known values:
ΔT = 80,000 J / (720.0 g * 4.18 J/g°C)
Calculating:
ΔT ≈ 26.82 °C
Therefore, the increase in temperature of the water will be approximately 26.82 degrees Celsius.