A 5.00 g sample of liquid water at 25.0 °C is heated by the addition of 145 J of energy. The final temperature of the water is ________ °C. The specific heat capacity of liquid water is 4.184 J g^-1 K^-1.

Question

Quizplus · Accepted Answer

We can use the formula:

q = mCΔT

Where: 
q = heat energy (145 J) 
m = mass of water (5.00 g) 
C = specific heat capacity of water (4.184 J/g°C) 
ΔT = change in temperature (we're solving for this)

Rearranging the formula:

ΔT = q / (mC) = 145 J / (5.00 g * 4.184 J/g°C) ≈ 6.94°C

Adding this change in temperature to the initial temperature of 25°C:

Final temperature = 25°C + 6.94°C ≈ 31.9°C

Therefore, the answer is (D) 31.9.

A 500 G Sample of Liquid Water at 25