High School

Calculate the energy required to heat 825.0 g of water from 26.0°C to 38.4 "C. Assume the specific heat capacity of water under these conditions is 4.18 J'g'K^-1

Answer :

Final answer:

The energy required to heat 825.0 g of water from 26.0°C to 38.4°C is 41,091 J.

Explanation:

To calculate the energy required to heat the water, we can use the formula:

Q = mcΔT

Where:

  • Q is the heat energy
  • m is the mass of the substance
  • c is the specific heat capacity
  • ΔT is the change in temperature

Given:

  • Mass of water (m) = 825.0 g
  • Specific heat capacity of water (c) = 4.18 J/g/K
  • Initial temperature (T1) = 26.0°C
  • Final temperature (T2) = 38.4°C

First, we need to calculate the change in temperature:

ΔT = T2 - T1 = 38.4°C - 26.0°C = 12.4°C

Now, we can substitute the values into the formula:

Q = (825.0 g) * (4.18 J/g/K) * (12.4°C)

Calculating the value:

Q = 41,091 J

Therefore, the energy required to heat 825.0 g of water from 26.0°C to 38.4°C is 41,091 J.

Learn more about calculating energy required to heat water here:

https://brainly.com/question/24694603

#SPJ14