Answered: A) water can be sterilized by boiling.…

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...

See similar textbooks

Related questions

Concept explainers

Question

A) water can be sterilized by boiling. How much heat is required to heat 4 gallons of water from room temperature 72F to 212F (1 g = 3.78 L)

B) if the heat of vaporization of water is 40.7 kJ/mol how much heat is required to convert the original 4 gallons of water not at 212F to steam. Assume the density of water is 1g/mL at 72F

Expert Solution

Step 1

Heat of Vaporization:

Heat of vaporization is the heat required to vaporize or evaporate to a certain amount of liquid to its gaseous form. It is represented by $∆ H_{v a p}$ . It is expressed in $k J m o l^{- 1}$

To calculate the heat required for any given amount of a substance the equation employed is-

$\begin{array}{rcl} Q & = & m \times c \times ∆ T \\ w h e r e, \\ Q & = & H e a t r e q u i r e d (J o r k J) \\ c & = & s p e c i f i c h e a t (\frac{J}{g ° C}) \\ ∆ T & = & D i f f e r e n c e i n t e m p e r a t u r e \\ ∆ T & = & F i n a l t e m p e r a t u r e (T_{f}) - I n i t i a l t e m p e r a t u r e (T_{i}) \end{array}$

Step 2

Answer to Part A:

Given,

$\begin{array}{rcl} V o l u m e o f w a t e r & = & 4 g a l l o n s \\ I n i t i a l T e m p e r a t u r e, T_{i} & = & 72 ° F \\ F i n a l T e m p e r a t u r e, T_{f} & = & 212 ° F \\ C o n v e r s i o n F a c t o r, 1 g a l l o n & = & 3.78 L \\ D e n s i t y o f w a t e r & = & 1 g / m L \end{array}$

First, convert gallons of water to Litres and then to millilitres

$1 g a l l o n = 3.78 L T h u s, 4 g a l l o n = 4 \times 3.78 L = 15.12 L N o w, 1 L = 1000 m L T h e n, 15.12 L = 15.12 \times 1000 m L = 15120 m L h e n c e, V o l u m e o f w a t e r, V = 15120 m L$

Next, use density to find the mass of water

$\begin{array}{rcl} G i v e n, \\ D e n s i t y & = & 1 g / m L \\ V o l u m e & = & 15120 m L \\ D e n s i t y & = & \frac{M a s s}{V o l u m e} \\ M a s s & = & D e n s i t y \times V o l u m e \\ M a s s & = & 1 \frac{g}{m L} \times 15120 m L \\ M a s s o f w a t e r & = & 15120 g \end{array}$

Step 3

Convert given temperatures from Fahrenheit to Celsius and find the temperature difference

$T h e f o r m u l a t o f i n d F a h r e n h e i t t o C e l s i u s i s - (33.8 ° F - 32) \times \frac{5}{9} = 0 ° C T_{i} = 72 ° F (72 ° F - 32) \times \frac{5}{9} = 22.22 ° C T_{i} = 22.22 ° C A l s o, T_{f} = 212 ° F (212 ° F - 32) \times \frac{5}{9} = 100 ° C T_{f} = 100 ° C T h u s, t e m p e r a t u r e d i f f e r e n c e, ∆ T = (100 - 22.22) ° C ∆ T = 77.78 ° C$

Step by step

Solved in 6 steps

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions