If 6.3KJ of energy is given to 300g of water, which is at 1⁰ C, what would be the final temperature?

To determine the final temperature of the water after adding 6.3 kJ of energy, we can use the specific heat capacity formula. This formula relates the amount of heat energy added to a substance, its mass, and the change in temperature. The specific heat capacity of water is approximately 4.18 J/g°C, which means it takes 4.18 joules to raise the temperature of 1 gram of water by 1 degree Celsius.

Understanding the Problem

We know the following:

• Energy added (Q) = 6.3 kJ = 6300 J (since 1 kJ = 1000 J)
• Mass of water (m) = 300 g
• Initial temperature (T_initial) = 10°C

Applying the Formula

The formula to calculate the change in temperature (ΔT) when heat is added is:

Q = m × c × ΔT

Where:

• Q = heat energy (in joules)
• m = mass of the substance (in grams)
• c = specific heat capacity (in J/g°C)
• ΔT = change in temperature (in °C)

Rearranging the Formula

We can rearrange the formula to find ΔT:

ΔT = Q / (m × c)

Calculating the Change in Temperature

Now, substituting the known values into the rearranged formula:

ΔT = 6300 J / (300 g × 4.18 J/g°C)

ΔT = 6300 J / 1254 J/°C

ΔT ≈ 5.02°C

Finding the Final Temperature

To find the final temperature (T_final), we add the change in temperature to the initial temperature:

T_final = T_initial + ΔT

T_final = 10°C + 5.02°C

T_final ≈ 15.02°C

Conclusion

After adding 6.3 kJ of energy to 300 grams of water initially at 10°C, the final temperature of the water would be approximately 15.02°C. This calculation illustrates how energy transfer affects temperature, showcasing the relationship between heat energy, mass, and temperature change in a straightforward manner.