How much heat must be removed from 25.0g of steam at 118.0C in order to form ice at 15C

-10778.95 J heat must be removed in order to form the ice at 15 °C.

Explanation:

Given dа ta:

mass of steam = 25 g

Initial temperature = 118 °C

Final temperature = 15 °C

Heat released = ?

Solution:

Formula:

q = m. c. ΔT

we know that specific heat of water is 4.186 J/g.°C

ΔT = final temperature - initial temperature

ΔT = 15 °C - 118 °C

ΔT = - 103 °C

now we will put the values in formula

q = m. c. ΔT

q = 25 g * 4.186 J/g.°C * - 103 °C

q = - 10778.95 J

so, - 10778.95 J heat must be removed in order to form the ice at 15 °C.