If 420 joules of heat energy is added to 25 grams of water at 25 degrees celcius, what will be the final temperature of the water, in Celcius degrees?

T2 = 29°C

Explanation:

Given dа ta:

Heat added = 420 j

Mass of water = 25 g

Initial temperature = 25°C

Final temperature = ?

Solution;

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water = 4.18 j/g.°C

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

Now we will put the values.

420 j = 25 g * 4.18 j/g.°C * (Final temperature - initial temperature)

420 j = 25 g * 4.18 j/g.°C * (T2 - 25°C)

420 j = 104.5 j/°C * (T2 - 25°C)

420 j / 104.5 j/°C = T2 - 25°C

4°C + 25°C = T2

T2 = 29°C