How much heat in kilojoules is needed to bring 1.49 kg of water from 28.1 to 97.1 °C (comparable to making four cups of coffee)?
q = mass water x specific heat water x (Tfinal-Tinitial)
To calculate the amount of heat needed to bring the water from 28.1 °C to 97.1 °C, we can use the specific heat equation: q = m * c * ΔT.
- q represents the amount of heat absorbed or released (in joules),
- m is the mass of the substance (in kilograms),
- c is the specific heat capacity of the substance (in joules per kilogram per degree Celsius),
- ΔT is the change in temperature (in degrees Celsius).
First, let's find the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 joules per gram per degree Celsius, or 4180 joules per kilogram per degree Celsius.
Now, let's calculate the amount of heat required using the equation:
q = m * c * ΔT
Given:
m (mass of water) = 1.49 kg
c (specific heat capacity of water) = 4180 J/kg°C
ΔT (change in temperature) = 97.1 °C - 28.1 °C = 69 °C
q = 1.49 kg * 4180 J/kg°C * 69 °C
q ≈ 445,869.6 joules
To convert the result to kilojoules, divide the value by 1000:
q ≈ 445.87 kilojoules
Therefore, approximately 445.87 kilojoules of heat would be needed to bring 1.49 kg of water from 28.1 °C to 97.1 °C.