how many calories are absorbed when 250.0 grams of warm water from 20.0c to 60.0c?
To calculate the amount of calories absorbed by warm water, we need to use the equation:
q = m * C * ΔT
Where:
q is the amount of heat energy absorbed (in calories)
m is the mass of the substance (in grams)
C is the specific heat capacity of the substance (in calories/gram°C)
ΔT is the change in temperature (in °C)
First, let's find the specific heat capacity of water. The specific heat capacity of water is 1.00 cal/g°C.
Now, we can calculate the amount of heat energy absorbed by using the given values:
m = 250.0 grams (mass of water)
ΔT = 60.0°C - 20.0°C = 40.0°C (change in temperature)
q = m * C * ΔT
q = 250.0 g * 1.00 cal/g°C * 40.0°C
Now we can calculate the value:
q = 10,000 calories
Therefore, 250.0 grams of warm water will absorb 10,000 calories when heated from 20.0°C to 60.0°C.