How much the heat needed to raise the temperature of 700 g of water from 25 c to 90 c?

2 Answers
Apr 20, 2018

Q=190372 J or
Q approx 190 KJ

Explanation:

To calculate the total heat needed to raise the temperature of 700g of water from 25 C to 90 C you would use the formula

Q=mcDeltaT

Where Q is heat (Joules), m is mass (grams), c is the specific heat of the substance (J/(g*C)), and DeltaT is the change in temperature (Celsius).

Specific heat of water, the energy/heat needed to raise the temperature of 1 gram of water by 1 degree Celsius is 4.184J/(g*C)

DeltaT is just the change in temperature 90C-25C=65C

Mass is 700g

From here you just put in each variable and solve for Q:

Q=700*4.184*65

Q=190372 J or
Q approx 190 KJ

Apr 20, 2018

Approximately "190.4 kJ"

Explanation:

Use this equation

"Q = mC"Δ"T"

where

  • "Q =" Heat
  • "m =" Mass of sample
  • "C =" Specific heat of sample ("4.184 J/g°C" for water)
  • "ΔT =" Change in temperature

"Q" = 700 cancel"g" × "4.184 J"/(cancel"g" cancel"°C") × (90 - 25) cancel"°C" = "190372 J" ≈ "190.4 kJ"