   Chapter 5, Problem 93IL

Chapter
Section
Textbook Problem

A piece of lead with a mass of 27.3 g was heated to 98.90 °C and then dropped into 15.0 g of water at 22.50 °C. The final temperature was 26.32 °C. Calculate the specific heat capacity of lead from these data.

Interpretation Introduction

Interpretation:

The specific heat capacity of lead 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 the lead is 27.3g

Specific heat capacity of water is 4.184J/gK

Assume qmetal=-qwater

Substitute the values in the equations:

q=C×m×ΔT

27.3g×Cmetal(299.47372.05)K

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