A cup that contains 5 ounces of water absorbs 1080J of energy from a heater. If

Question

A cup that contains 5 ounces of water absorbs 1080J of energy from a heater. If the intial temperature of the water was 20 C, what is the final temperature of the water?

Answer sheet says final temperature (T2) is 21.8 C

but I am unsure how. My Calculations are assume it includes the Cp of Water=4180J/kg*C

(0.1417475kg)(4180J/kg*c)(20 C) = 1080J x T2

which gives 11850.091J / 1080 J = 10.97 C

Final answer is 10.97C but 10.97 C=/= Correct answer of 21.8 C. Seeking guidance on what step I missed and the thought process behind the answer.

Explanation / Answer

Here, 1 ounce = 28.35 gm

Mass (m)= 5 x 28.35= 141.75 gm

Specific heat (c) = 4.18 J/ gm- degree cel.

Heat (Q) = mc (T2 - T1)

T2 - T1 = Q/mc

T2 = T1 + Q/mc

= 20 + 1080/141.75*4.18

= 20 + 1.8= 21.8 C

T2 = 21.8 C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.