Expert solution/answer Specific heat of the water is 4.18 kJ/kg. °C. The expression for the heat transfer of the cold beverage is: Q = mc(T, -T) cold i cold 10 m (4.18 kJ/kg - °C)(48°C –16°C) 1 mL = (120 mL ) (10 J = (16.0×10* kJ) 1 kJ = 16 J I have used the expert formula above in calculating (d), and it happens to be section (c) answer is wrong but (d), is correct...may you please re-check for me where the problem is with expert's solution (c.) Solution (d) Q = MhotC (Tf – T;i hot) 16 J I have used the expert formula above in calculating (d), and it happens to be section (c) answer is wrong but (d), is correct...may you please re-check for me where the problem is with expert's solution (c.) Solution (d) Q = MhotC (Tf – Ti hot) [10-5m²- (120mL) (4. 18 kJ/kg.°C) (48-69°C) 1ml (103 ('10.5 x 10*kJ) () kJ = 10.5 Joules The formula above has worked for section (d)...I don't know why section c is not correct.

Inquiry into Physics
8th Edition
ISBN:9781337515863
Author:Ostdiek
Publisher:Ostdiek
Chapter5: Temperature And Heat
Section: Chapter Questions
Problem 13P: - (a) Compute the amount of heat needed to raise the temperature of 1 kg of water from its freezing...
icon
Related questions
icon
Concept explainers
Question
100%

Initial temperature hot water 69° C,
Final temperature after adding cold water 48° C.
Initial temperature of cold water is 16° C 
Quantity warm water 175 ml

c)What is your calculated heat transfer (Q) of the cold beverage when 4 ounces (120 mL) was added to the hot beverage? Report your answer in Joules.

d) What is your calculated heat transfer (Q) of the hot beverage when 4 ounces (120 mL) of cold beverage was added? Report your answer in Joules.

 

Expert solution/answer
Specific heat of the water is 4.18 kJ/kg. °C.
The expression for the heat transfer of the cold beverage is:
Q = mc(T, -T)
cold
i cold
10 m
(4.18 kJ/kg - °C)(48°C –16°C)
1 mL
= (120 mL )
(10 J
= (16.0×10* kJ)
1 kJ
= 16 J
I have used the expert formula above in calculating (d), and it happens to be section
(c) answer is wrong but (d), is correct...may you please re-check for me where the
problem is with expert's solution (c.)
Solution (d)
Q = MhotC (Tf – T;i hot)
Transcribed Image Text:Expert solution/answer Specific heat of the water is 4.18 kJ/kg. °C. The expression for the heat transfer of the cold beverage is: Q = mc(T, -T) cold i cold 10 m (4.18 kJ/kg - °C)(48°C –16°C) 1 mL = (120 mL ) (10 J = (16.0×10* kJ) 1 kJ = 16 J I have used the expert formula above in calculating (d), and it happens to be section (c) answer is wrong but (d), is correct...may you please re-check for me where the problem is with expert's solution (c.) Solution (d) Q = MhotC (Tf – T;i hot)
16 J
I have used the expert formula above in calculating (d), and it happens to be section
(c) answer is wrong but (d), is correct...may you please re-check for me where the
problem is with expert's solution (c.)
Solution (d)
Q = MhotC (Tf – Ti hot)
[10-5m²-
(120mL)
(4. 18 kJ/kg.°C) (48-69°C)
1ml
(103
('10.5 x 10*kJ) ()
kJ
= 10.5 Joules
The formula above has worked for section (d)...I don't know
why section
c is not correct.
Transcribed Image Text:16 J I have used the expert formula above in calculating (d), and it happens to be section (c) answer is wrong but (d), is correct...may you please re-check for me where the problem is with expert's solution (c.) Solution (d) Q = MhotC (Tf – Ti hot) [10-5m²- (120mL) (4. 18 kJ/kg.°C) (48-69°C) 1ml (103 ('10.5 x 10*kJ) () kJ = 10.5 Joules The formula above has worked for section (d)...I don't know why section c is not correct.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 2 images

Blurred answer
Knowledge Booster
Energy transfer
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Inquiry into Physics
Inquiry into Physics
Physics
ISBN:
9781337515863
Author:
Ostdiek
Publisher:
Cengage
Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
Physics
ISBN:
9781133104261
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
College Physics
College Physics
Physics
ISBN:
9781938168000
Author:
Paul Peter Urone, Roger Hinrichs
Publisher:
OpenStax College
University Physics Volume 2
University Physics Volume 2
Physics
ISBN:
9781938168161
Author:
OpenStax
Publisher:
OpenStax
College Physics
College Physics
Physics
ISBN:
9781285737027
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
College Physics
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning