24.1 kJ of energy is lost by a 251 g Al block. If the block is initially at 125.0°C what will be its final T in °C? (sAl = 0.902 J g-1 °C-1)I need help on what i did wrong on the problem, the problem is on the picture and the second picture is how I did it, please tell me what I did wrong, thank you A 24.1 kJ of energy is lost by a 251 g aluminum block. If the block isinitially at 125.0°C what will be its final temperature? (CA = 0.902 Jg1°C1)´Al -24.1(1000)=251(0.902)(Tf-125C)-24100=2264.02(Tf) - 125(2264.02)-24100+125 (2264.02)=2264.02(Tf)258902.5=2264.02TfTf=114 Celcius

Answered: Image /qna-images/answer/19e2408d-2e3c-4e90-9ead-849c05408df0.jpg

Answered: final T

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Concept explainers

Question

100%

24.1 kJ of energy is lost by a 251 g Al block. If the block is initially at 125.0°C what will be its final T in °C? (s_Al = 0.902 J g-1 °C-1)

I need help on what i did wrong on the problem, the problem is on the picture and the second picture is how I did it, please tell me what I did wrong, thank you

A 24.1 kJ of energy is lost by a 251 g aluminum block. If the block is
initially at 125.0°C what will be its final temperature? (CA = 0.902 J
g1°C1)
´Al

-24.1(1000)=251(0.902)(Tf-125C)
-24100=2264.02(Tf) - 125(2264.02)
-24100+125 (2264.02)=2264.02(Tf)
258902.5=2264.02Tf
Tf=114 Celcius

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions