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
babunello [35]
3 years ago
5

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi

nutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.
A) 1900 J/kg K
B) 2270 J/kg. K
C) 3300 J/kg K
D) 3800 J/kg K
E) 4280 J/kg K
Physics
1 answer:
Nataliya [291]3 years ago
7 0

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

m_{m} = mass of metal = 400 g

c_{m} = specific heat of metal = ?

T_{mi} = initial temperature of metal = 100 °C

m_{a} = mass of aluminum cup = 100 g

c_{a} = specific heat of aluminum cup = 900.0 J/kg ∙ K

T_{ai} = initial temperature of aluminum cup = 15 °C

m_{w} = mass of water = 500 g

c_{w} = specific heat of water = 4186 J/kg ∙ K

T_{wi} = initial temperature of water = 15 °C

T = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}

You might be interested in
A labor push a crate through 12m and 720 J of work .The magnitude of force was
Harrizon [31]

Answer:

<h2>60 N</h2>

Explanation:

The magnitude of the force can be found by using the formula

f =  \frac{w}{d}  \\

w is the workdone

d is the distance

From the question we have

f =  \frac{720}{12}  = 60 \\

We have the final answer as

<h3>60 N</h3>

Hope this helps you

6 0
2 years ago
Compared to the tropical rainforests, the temperate rainforests generally have __________.
Oxana [17]
<span>a. less biomass
b. more biomass
c. more broad-leaf trees
d. locations near the equator


i would say c
</span>
8 0
3 years ago
When energy is transferred to air, what happens to the particles of air? (1 point) They move slower. They move faster. They cool
zubka84 [21]
This question is a critical question. as we all know, when energy is added to any state of water, the particles move faster. and when energy is taken away from any state of water, the particles reduce speed. same with the particles of air. when energy is added; they move faster. when energy is removed; they move slower. so the answer is they move faster
5 0
3 years ago
Read 2 more answers
A stone dropped in a pond sends out a circular ripple whose radius increases at a constant rate of 4 ft/sec. After 12 seconds, h
Natali5045456 [20]

Answer:

After 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

Explanation:

Area of a circle = πr²

where;

r is the circle radius

Differentiate the area with respect to time.

\frac{dA}{dt} = 2\pi r\frac{dr}{dt}

dr/dt = 4 ft/sec

after 12 seconds, the radius becomes = \frac{dr}{dt} X 12 = 4 \frac{ft}{sec}  X 12 sec = 48 ft

To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt

\frac{dA}{dt} = 2\pi r\frac{dr}{dt}

\frac{dA}{dt} = 2\pi (48)(4)

    dA/dt = 1206.528 ft²/sec

Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

7 0
3 years ago
Is the expression "The bigger they are, the harder they fall" a generally true statement since, in the absence of air resistance
lions [1.4K]
No, because in oxygen depraved rooms, if you drop a feather and a bowling ball at the same height and time, they will fall at the same speed and have the same amount of impact.
6 0
3 years ago
Read 2 more answers
Other questions:
  • When writing a lab report where do your data tables belong?
    15·2 answers
  • What are gamma rays used for
    15·2 answers
  • Which object has potential energy but not kinetic energy
    14·1 answer
  • Whích best explains parallel forces?
    9·1 answer
  • A proton is suspended in the air by an electric field at the surface of Earth. What is the strength of this electric field?
    9·1 answer
  • The number of electrons with the number of protons in a charged object
    8·1 answer
  • What are 3 common sources of voltage difference?
    5·1 answer
  • What<br>is the CGS and<br>SI unit of weight?​
    13·2 answers
  • Calculate the current through a 500W,230V heater. Which fuse would you use for this appliance?
    5·1 answer
  • 5. Find the velocity of a train that traveled 75 km in 35 minutes. (answer
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!