1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Mnenie [13.5K]
3 years ago
10

If an 83.00 g sample of Iron has a starting temperature of 297K and an ending temperature of 329K how much heat will be lost fro

m the Iron samples?
If the heat absorbed from a 59.00 g sample is 4814.4 J and the temperature change is 85.0C, what is the samples specific heat? What is the sample?

The heat absorbed by a sample of 55.00 g is 3908.3 J and its starting temperature is 78.0C and its ending temperature is 95.0C, what is the specific heat of this sample and what is the sample?
Physics
1 answer:
diamong [38]3 years ago
4 0

Question 1 :

m = mass of iron sample = 83

c = specific heat of iron sample = 0.450 J/g ⁰C

T₀ = initial temperature = 297 K

T = final temperature = 329 K

Q = heat lost

Heat lost is given as

Q = m c (T - T₀)

inserting the values

Q = (83) (0.450) (329 - 297) = 1195.2 J



Question 2 :

m = mass of sample = 59 g

c = specific heat of sample = ?

ΔT = change in temperature = 85 c

c = specific heat

Q = heat absorbed = 4814.4 J

Heat absorbed is given as

Q = m c

inserting the values

4814.4 = (59) (85) c

c = 0.96 J/g C

the sample is cholorofom


Question 3 :


m = mass of sample = 55 g

c = specific heat of sample = ?

T₀ = initial temperature = 78 c

T = final temperature = 95 c

Q = heat absorbed = 3908.3 J

Heat absorbed is given as

Q = m c (T - T₀)

inserting the values

3908.3 = (55) (95 - 78) c

c = 4.2 J/g C

the sample is water



You might be interested in
Does the sign of the charge (positive or negative) affect how that charge is dissipated? Explain.
labwork [276]
This is an add proceeded by
5 0
3 years ago
Please help meh its due
Fed [463]
-0.000393025 light years
3 0
2 years ago
Read 2 more answers
A 1.15 kg book is at rest on the table. What is the magnitude of the normal force that the table is exerting on the book?
Agata [3.3K]

Answer:

11.27N

Explanation:

Given parameters:

Mass of the book  = 1.15kg

Unknown:

Magnitude of the normal force  = ?

Solution:

The normal force is the vertical force exerted by a body on an object.

It can be described as the weight of an object.

 Normal force  = mass x acceleration due to gravity

 Normal force  = 1.15 x 9.8  = 11.27N

5 0
2 years ago
One difference between a hypothesis and a theory is that a hypothesis
artcher [175]
A hypothesis is a tentative statement which is made to try to explain a known phenomenon but whose truth value is still uncertain, whether it is true or no depends on further research. On the other hand, a theory is made up of hypothesis which have been proven to be true so far, a theory should be able to explain future phenomena successfully
3 0
3 years ago
What is the acceleration of a 10 kg mass pushed by a 5 N force
hichkok12 [17]

Answer:

The formula is a = F m so in this case a = 5 10 = 0.5 m s 2

Explanation:

3 0
2 years ago
Other questions:
  • Which two biomes have little precipitation?
    8·1 answer
  • I NEED AN ANSWER ASAP!!!!!!!!!!!! PLEASE HELP ME!!!!!!!!!!!!!!! A 1.5m wire carries a 9 A current when a potential difference of
    6·1 answer
  • How is the kelvin scale different from the Celsius scale?
    6·1 answer
  • What hormone is not related to stress
    7·1 answer
  • Does a wave travel through solid, liquid, or gas?
    9·2 answers
  • How does Newton's first law of motion affect people in a car crash?​
    11·1 answer
  • There are three known forms of uranium. These forms are called _____ of each other.
    10·1 answer
  • Pls answer ....................................
    5·1 answer
  • What is the elevation at point A?
    6·1 answer
  • if your oven uses a 220.0 volt line and draws a maximum 8.00 A current what is the resistance of the oven when it is fully heate
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!