How much heat is absorbed when 500. g of water Cp 4.184 Jg C goes from 25.0 C to 35.0 C?

To calculate the heat absorbed by the water, use the formula ( q = m \cdot C_p \cdot \Delta T ), where ( q ) is the heat absorbed, ( m ) is the mass of the water (500 g), ( C_p ) is the specific heat capacity (4.184 J/g°C), and ( \Delta T ) is the change in temperature (35.0°C - 25.0°C = 10.0°C).

Plugging in the values:

[ q = 500 , \text{g} \times 4.184 , \text{J/g°C} \times 10.0 , \text{°C} = 20920 , \text{J} ]

Therefore, the heat absorbed is 20,920 J.