calculate the energy in joules lost when 70.0 grams of water cools from 86.4 celsius to 7.1 celsius
joules lost when 29 g of water cools from 86∘C to 69 ∘C
To calculate the energy lost when water cools from one temperature to another, we need to use the specific heat capacity formula and the equation for heat transfer.
The specific heat capacity (c) of water is 4.18 J/g°C. This means that it takes 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
First, we need to calculate the change in temperature:
ΔT = final temperature - initial temperature
ΔT = 7.1°C - 86.4°C
ΔT = -79.3°C
Since we are calculating the energy lost, the change in temperature is negative.
Next, we multiply the change in temperature by the mass of water and the specific heat capacity to find the energy lost:
Energy Lost = mass × specific heat capacity × ΔT
Energy Lost = 70.0 g × 4.18 J/g°C × -79.3°C
Simplifying the calculation:
Energy Lost = -23379.4 J
The energy lost when 70.0 grams of water cools from 86.4°C to 7.1°C is approximately -23379.4 Joules. The negative sign indicates that energy is being lost.