Final answer:
Using the specific heat capacity of water and the formula for thermal energy transfer, the energy required to warm a 150.0g sample of water from 30.00°C to 63.50°C is found to be 21000 J.
Step-by-step explanation:
To determine how much energy was required to warm the sample of water from 30.00 degrees Celsius to 63.50 degrees Celsius, we can use the formula for thermal energy transfer (Q):
Q = mcΔT
where:
- Q is the thermal energy transfer
- m is the mass of the water
- c is the specific heat capacity of water
- ΔT is the change in temperature
The specific heatof water (c) is 4.184 J/g°C
. The mass of the water (m) is 150.0 g, and the temperature change (ΔT) is 33.50°C (63.50°C - 30.00°C).
Using the formula:
Q = (150.0 g)(4.184 J/g°C)(33.50°C)
Q = 21000 J
Therefore, the energy required to warm the sample of water is 21000 J.