Answered

Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

What is the temperature increase of 4.0 kg of water when it is heated by an 8.0 ^102 W immersion heater for exactly 10.0 min? (cp = 4186 J/kg·°C)

Sagot :

samuelpoly011

Answer:

ΔT = 28.667°C is the answer.

Explanation:

Given,

Mass of water = 4.0 kg

Water is heated by = 8 × 10² W = 800 W

Time = 10 min = 10 × 60 = 600 s

Now,

using the equation of specific heat,

Q = mcΔT

where c is the specific heat capacity of water = 4186 J/kg°C

Q = Pt

Substituting,

Pt = mcΔT

800 × 600 = 4 × 4186 × ΔT

480000 = 16744 × ΔT

ΔT = 28.667

∴Temperature increased, ΔT = 28.667°C

We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.