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 $4.184 \mathrm {~J} \mathrm {~g} ^ { - 1 } \mathrm {~K} ^ { - 1 }$

Question

Quizplus · Accepted Answer

We can use the formula 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 this equation we get:
ΔT = q / (mC)
Plugging in the values we get:
ΔT = 145 J / (5.00 g x 4.184 J/g°C)
ΔT = 6.914°C
Adding this to the initial temperature of 25.0°C, we get the final temperature of:
25.0°C + 6.914°C = 31.914°C ≈ 31.9°C, which is closest to answer choice E. Therefore, the answer is E.

A 500 G Sample of Liquid Water at 25 4.184 J g−1 K−14.184 \mathrm {~J} \mathrm {~g} ^ { - 1 } \mathrm {~K} ^ { - 1 }4.184 J g−1 K−1

A 500 G Sample of Liquid Water at 25 $4.184 \mathrm {~J} \mathrm {~g} ^ { - 1 } \mathrm {~K} ^ { - 1 }$