When 3.0 kg of water is cooled from 80.0C to 10.0C, how much heat energy is lost?

the formula is 4.179

To determine the heat energy lost when water is cooled, we can use the heat energy equation:

Q = mcΔT

Where:
Q is the heat energy lost
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature

First, let's calculate the change in temperature:

ΔT = Final temperature - Initial temperature
ΔT = 10.0°C - 80.0°C
ΔT = -70.0°C

Since the water is being cooled, the change in temperature is negative.

Next, we need to find the specific heat capacity of water. The specific heat capacity of water is 4.18 J/g·°C.

Now we need to convert the mass of the water from kilograms to grams:

m = 3.0 kg * 1000 g/kg
m = 3000 g

Now we can calculate the heat energy lost:

Q = mcΔT
Q = (3000 g) * (4.18 J/g·°C) * (-70.0°C)

Calculating this equation will give us the amount of heat energy lost.

Are you looking for

q = mass H2O x specific heat H2O x (Tfinal - Tintial)

don't you have some handy formula for water with units of J/°K ?