How much heat is required to warm 1.60 L of water from 26.0 degree C to 100.0 degree C? (Assume a density of {1.0 g/ml for the water.)
q = mass water x specific heat water x (Tfinal-Tinitial)
118400
To calculate the amount of heat required to warm the water, we can use the formula:
Q = mcΔT
Where:
Q = heat energy (in joules)
m = mass of the water (in grams)
c = specific heat capacity of water (4.18 J/g·°C)
ΔT = change in temperature (in °C)
First, let's calculate the mass of the water. Since the density of water is given as 1.0 g/ml, we can convert the volume of water to grams:
Mass of water = volume × density
Mass of water = 1.60 L × 1000 g/L (since 1 L = 1000 ml)
Mass of water = 1600 grams
Next, we can calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 100.0 °C - 26.0 °C
ΔT = 74.0 °C
Now, we can substitute these values into the formula to find the heat energy:
Q = mcΔT
Q = 1600 g × 4.18 J/g·°C × 74.0 °C
Q = 495,392 J (or 4.95 × 10^5 J)
Therefore, approximately 495,392 joules of heat energy is required to warm 1.60 L of water from 26.0 °C to 100.0 °C.