Answered

At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

5 x 10 ^ 4 * J of heat are added 35 kg of water initially at 12 degrees C What is the water's final temperature?

Sagot :

Answer:

1 cal / gm-C = 1000 cal /kgm-C      specific heat of water

4.19 joules = 1 C

5.0E4 J / 4.19 J/C = 1.2E4 Calories = ΔQ

ΔQ = K M (T2 - T1)       where K is the heat capacity

T2 = 12 + 1.2E4 / (1000 * 35)

T2 = 12 + 1.2E4 / 3.5 E4 = 12.34 C    (only .34 deg C)

We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.