Final answer:
Using the formula Q = m × c × ΔT and the specific heat capacity of water, it is found that 21,961.26 joules are released when 70.5g of water is cooled from 100.0°C to 25.5°C.
Step-by-step explanation:
To calculate the number of joules released when 70.5g of water is cooled from 100.0°C to 25.5°C, we will use the specific heat capacity of water and the formula for heat transfer:
Q = m × c × ΔT
where Q is the heat transferred, m is the mass of the water, c is the specific heat capacity of water (4.184 J/g°C), and ΔT is the change in temperature. Plugging in the values:
Q = 70.5g × 4.184 J/g°C × (25.5°C - 100.0°C)
Q = 70.5g × 4.184 J/g°C × (-74.5°C)
Q = -21961.26 J
Therefore, when 70.5g of water is cooled from 100.0°C to 25.5°C, 21,961.26 joules of heat are released.