   Chapter 5, Problem 13PS

Chapter
Section
Textbook Problem

A 182-g sample of gold at some temperature was added to 22.1 g of water. The initial water temperature was 25.0 °C, and the final temperature was 27.5 °C. If the specific heat capacity of gold is 0.128 J/g · K, what was the initial temperature of the gold sample?

Interpretation Introduction

Interpretation:

Initial temperature of gold mixed into water different temperature 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,

Mass of water =22.1g

For water,

Tfinal=27.50C

Tinitial=250C

Mass of gold =182g

Specific heat capacity of gold=0.128JK/g

Assume the sum of qwater  and qgold =0

Final temperature can be calculated as,

[Cwater×Mwater(Tfinal

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