   # A cubic piece of uranium metal (specific heat capacity = 0.117 J/ ° C • g) at 200.0°C is dropped into 1.00 L deuterium oxide ("heavy water;" specific heat capacity = 4.21 I J/° C • g) at 25.5°C. The final temperature of the uranium and deuterium oxide mixture is 28.5°C. Given the densities of uranium (19.05 g/cm 3 ) and deuterium oxide (1.11 g/mL). what is the edge length of the cube of uranium? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 7, Problem 133IP
Textbook Problem
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## A cubic piece of uranium metal (specific heat capacity = 0.117 J/°C • g) at 200.0°C is dropped into 1.00 L deuterium oxide ("heavy water;" specific heat capacity = 4.21 I J/°C • g) at 25.5°C. The final temperature of the uranium and deuterium oxide mixture is 28.5°C. Given the densities of uranium (19.05 g/cm3) and deuterium oxide (1.11 g/mL). what is the edge length of the cube of uranium?

Interpretation Introduction

Interpretation: The edge length of the cube of uranium should be determined.

Concept Introduction:

The heat capacity C is defined as the relation of heat absorbed to the temperature change. It can be given by,

C = Heat absorbedTemperature change......(1)

Require heat for an one gram of substance raise to its temperature by one degree Celsius is called specific heat capacity.

Absorbed heat (J)=Specific heat capacity×Temperature change(c)×mass ofsubstance (g)......(2)

For the above equation heat is:

S = q×M×T.....(1)

q is heat (J)

M is mass of sample (g)

S is specific heat capacity (J/°C·g.)

T is temperature change (C)

For the process no heat loss to the surroundings means then the heat is

(absorbed)-q×M×ΔT=-q×M×ΔT(released)......(3)

### Explanation of Solution

Explanation

Given data:

Specific heat capacity of uranium = 0.117 J/°C· g

Specific heat capacity of heavy water = 4.18 J/°C · g

Specific heat capacity of steam = 2.02J/°C· g

Volume of heavy  water is 1.0L

Initial temperature of uranium is 200°C

Final temperature of the mixture is 28.5°C.

Density of uranium is 19.05 g/cm3Densityofheavywateris1.11 g/mL

To calculate the required mass of heavy water.

Massofheavywater=volume×density=1.00×103mL×1.11gmL=1110g

• The volume of heavy water is multiplied by the density to give the mass of heavy water.
• The mass of heavy water is 1110g.

To calculate the heavy water gained heat.

q(gain)=4.211J°C.g×1110g×(28.525.2)°C=1.4×104J

• The calculated mass of heavy water, given specific heat capacity and temperature change are plugging in to equation (1) to give a heat gain by heavy water.
• The heat gain by heavy water is 1.4×104J.

To calculate the mass of uranium.

(absorbed)-qHeavywater×M×ΔT=-quranium×M×ΔT(released)

Heatlossuranium=0

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