   Chapter 5, Problem 12PS

Chapter
Section
Textbook Problem

One beaker contains 156 g of water at 22 °C, and a second beaker contains 85.2 g of water at 95 °C. The water in the two beakers is mixed. What is the final water temperature?

Interpretation Introduction

Interpretation:

The final temperature for water mixed at two different temperatures has to be calculated.

Concept Introduction:

Heat energy required to raise the temperature of 1g of substance by 1k.Energy gained or lost can be calculated using the below equation.

q=C×m×ΔT

Where,

q= energy gained or lost for a given mass of substance (m),

C =specific heat capacity

ΔT= change in temperature

Explanation

Given

At 220C , mass =156g

At 950C , mass =85.2g

Assume the sum of qwater at 293K  and qwater at 368K =0

Final temperature can be calculated as,

[Cwater×Mwater(Tfinal-T

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