Answered: 40.036 g of a metal is heated to 99.6…

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...

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Question

40.036 g of a metal is heated to 99.6 °C and is then placed into 25.0 g of water at 22.4 °C. The mixture reaches a temperature of 33.7 °C. What is the specific heat capacity of the metal?

Expert Solution

Step 1

Given:

$\begin{array}{rcl} Mass of metal & = & 40.036 g \\ Temperature of metal & = & 99.6 ° C \\ Mass of water & = & 25.0 g \\ Temperature of water & = & 22.4 ° C \\ Final temperature of mixture & = & 33.7 ° C \end{array}$

Step 2

To find the specific heat capacity of metal as follows,

The heat lost by metal is equal to the amount of heat absorbed by water.

$Metal Water m = 40.036 g m = 25.0 g T_{i} = 99.6 ° C T_{i} = 22.4 ° C T_{f} = 33.7 ° C c = 4.184 J / g . ° C$

Substituting the values in the following expression,

$\begin{array}{rcl} Heat lost by metal & = & Heat absorbed by water \\ q_{1} & = & q_{2} \\ {(m . c . ∆ T)}_{metal} & = & {(m . c . ∆ T)}_{water} \\ {(m . c . (T_{i} - T_{f}))}_{metal} & = & {(m . c . (T_{f} - T_{i}))}_{water} \\ (40.036 g) (c_{metal}) (99.6 - 33.7) ° C & = & (25.0 g) (4.184 J / g . ° C) (33.7 - 22.4) ° C \\ (2638.372 g ° C) (c_{metal}) & = & (1181.98 J) \\ c_{metal} & = & \frac{(1181.98 J)}{(2638.372 g ° C)} \\ c_{metal} & = & 0.448 J / g . ° C \end{array}$

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