If 720.0 mL of liquid water absorbs 80.0 kJ of heat, what will be its increase in temperature?
11th grade can not be the School Subject. Could we please begin there?
Sra
what grade do you work on geography?
To find the increase in temperature of the water, we need to use the equation:
q = m * C * ΔT
Where:
q = heat absorbed or released (in joules)
m = mass of the substance (in grams)
C = specific heat capacity of the substance (in J/g·°C or J/g·K)
ΔT = change in temperature (in °C or K)
First, let's convert the volume of water into its mass. Since we know the density of water is 1g/mL, the mass of 720.0 mL of water will be:
mass = volume * density = 720.0 mL * 1 g/mL = 720.0 g
Now we need to determine the specific heat capacity of water, which is approximately 4.18 J/g·°C.
Now we can rearrange the equation and solve for ΔT:
ΔT = q / (m * C)
ΔT = 80.0 kJ = 80.0 * 1000 J / (720.0 g * 4.18 J/g·°C)
ΔT = 80000 J / (720.0 g * 4.18 J/g·°C)
Plugging in the values and calculating:
ΔT ≈ 29.95 °C
Therefore, the increase in temperature of the water will be approximately 29.95 degrees Celsius.