How much heat is required to warm 1.60 L of water from 22.0 ∘C to 100.0 ∘C? (Assume a density of 1.0g/mL for the water.)
Express your answer in joules to two significant figures.
the specific heat of water is ... 4184 J per kg per ºC
4184 * 1.60 * (100.0 - 22.0) = ? J
To calculate the amount of heat required to warm water, we can use the formula:
q = m * c * ΔT
Where:
q is the heat required (in joules),
m is the mass of water (in grams),
c is the specific heat capacity of water (in J/g⋅°C), and
ΔT is the change in temperature (in °C).
First, we need to find the mass of water. Given that the density of water is 1.0 g/mL and the volume is 1.60 L, we can calculate the mass as:
mass = density * volume
mass = 1.0 g/mL * 1.60 L
mass = 1.60 kg
Next, we need to find the change in temperature. This can be calculated as:
ΔT = final temperature - initial temperature
ΔT = 100.0 °C - 22.0 °C
ΔT = 78.0 °C
Now, we need to find the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g⋅°C.
Now, we can substitute the values into the formula:
q = (1.60 kg) * (4.18 J/g⋅°C) * (78.0 °C)
q ≈ 522.86 J
Therefore, the amount of heat required to warm 1.60 L of water from 22.0 °C to 100.0 °C is approximately 522.86 joules.
To determine the heat required to warm the water, we can use the following equation:
q = m * c * ΔT
Where:
q is the heat (in joules)
m is the mass of the water (in grams)
c is the specific heat capacity of water (in joules/gram·degree Celsius)
ΔT is the change in temperature (in degrees Celsius)
First, we need to find the mass of the water using its volume and density. The density of water is given as 1.0 g/mL, which means that 1.60 L of water has a mass of:
mass = volume * density = 1.60 L * 1.0 g/mL = 1.60 kg = 1600 g
Next, we need to calculate the change in temperature:
ΔT = final temperature - initial temperature = 100.0 °C - 22.0 °C = 78.0 °C
Now we can substitute the values into the equation:
q = 1600 g * c * 78.0 °C
The specific heat capacity of water is approximately 4.18 J/g·°C. Substituting this value:
q = 1600 g * 4.18 J/g·°C * 78.0 °C
Calculating this expression will give us the heat required to warm the water.