Final answer:
To raise the temperature of 65.0 grams of water from 25.0 °C to 30.0 °C, 1,359.5 Joules of heat are required, calculated using the specific heat capacity of water and the mass and temperature change of the water.
Step-by-step explanation:
To calculate how much heat is needed to raise the temperature of 65.0 grams of liquid water from 25.0 degrees Celsius to 30.0 degrees Celsius, we can use the formula:
Q = mcΔT
Where Q is the heat transfer, m is the mass of the water, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature.
Let's plug in the values:
- m = 65.0 g
- c = 4.18 J/g°C
- ΔT = (30.0 °C - 25.0 °C) = 5.0 °C
Q = (65.0 g)(4.18 J/g°C)(5.0 °C)
Q = 65.0 * 4.18 * 5.0
Q = 1,359.5 J
So, 1,359.5 Joules of heat are required to raise the temperature of 65.0 grams of water from 25.0 °C to 30.0 °C.