Question

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?

Answer 1

T2 = 29°C

Step-by-step explanation:

Given data:

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