How much heat is required to warm 1.30 L of water from 30.0 degrees C to 100.0 degrees C?
(Assume a density of 1.0 g/mL for the water.)
292.88
To find the amount of heat required to warm the water, we can use the formula:
Q = m * c * ΔT
Where:
Q is the amount of heat
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
To find the mass of the water, we multiply the volume by the density:
m = V * ρ = (1.30 L) * (1.0 g/mL)
Next, we need to calculate the change in temperature:
ΔT = Final temperature - Initial temperature
ΔT = 100.0°C - 30.0°C
Now we can substitute the values into the formula:
Q = m * c * ΔT
Q = (1.30 L) * (1.0 g/mL) * c * (100.0°C - 30.0°C)
Given that the specific heat capacity of water is approximately 4.18 J/g·°C, we substitute this value to calculate Q:
Q = (1.30 L) * (1.0 g/mL) * (4.18 J/g·°C) * (100.0°C - 30.0°C)
Now we can multiply and simplify:
Q = (1.30 L) * (1.0 g/mL) * (4.18 J/g·°C) * (70.0°C)
To get the final answer, we multiply the values:
Q = 382.88 J
Therefore, it would require approximately 382.88 Joules of heat to warm 1.30 L of water from 30.0°C to 100.0°C.