A 4.50-g sample of liquid water at 25.0 °C is heated by the addition of 133 J of energy. The final temperature of the water is ________ °C. The specific heat capacity of liquid water is 4.18 J/g-K.
A)149
B)25.1
C)-17.9
D)32.1
E)7.07

Question

Quizplus · Accepted Answer

First, we need to use the equation:

q = mcΔT

where q is the energy added, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.

Rearranging the equation to solve for ΔT:

ΔT = q / mc

Plugging in the given values:

m = 4.50 g

c = 4.18 J/g-K

q = 133 J

ΔT = (133 J) / (4.50 g x 4.18 J/g-K) = 7.07 K

Since the initial temperature was 25.0 °C, the final temperature will be:

Final temperature = Initial temperature + ΔT = 25.0 °C + 7.07 K = 32.1 °C

Therefore, the best choice is D.

A 450-G Sample of Liquid Water at 25