If 250.0 g of water at 30.0 °C cool to 5.0 °C, how many kilojoules of energy did the water lose?

-26.125 kj

Explanation:

Given dа ta:

Mass of water = 250.0 g

Initial temperature = 30.0°C

Final temperature = 5.0°C

Amount of energy lost = ?

Solution:

Formula:

Q = m. c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 5.0°C - 30.0°C

ΔT = - 25°C

Specific heat of water is 4.18 j/g.°C

Now we will put the values in formula.

Q = m. c. ΔT

Q = 250.0 g * 4.18 j/g.°C * - 25°C

Q = - 26125 j

J to kJ

-26125 j * 1 kj / 1000 j

-26.125 kj