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
tekilochka [14]
3 years ago
10

Find the quantity of heat needed

Physics
1 answer:
krok68 [10]3 years ago
3 0

Answer:

Approximately 3.99\times 10^{4}\; \rm J (assuming that the melting point of ice is 0\; \rm ^\circ C.)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}

The energy required comes in three parts:

  • Energy required to raise the temperature of that 0.100\; \rm kg of ice from (-10\; \rm ^\circ C) to 0\; \rm ^\circ C (the melting point of ice.)
  • Energy required to turn 0.100\; \rm kg of ice into water while temperature stayed constant.
  • Energy required to raise the temperature of that newly-formed 0.100\; \rm kg of water from 0\; \rm ^\circ C to 10\;\ rm ^\circ C.

The following equation gives the amount of energy Q required to raise the temperature of a sample of mass m and specific heat capacity c by \Delta T:

Q = c \cdot m \cdot \Delta T,

where

  • c is the specific heat capacity of the material,
  • m is the mass of the sample, and
  • \Delta T is the change in the temperature of this sample.

For the first part of energy input, c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}.

Similarly, for the third part of energy input, c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}.

The second part of energy input requires a different equation. The energy Q required to melt a sample of mass m and latent heat of fusion L_\text{f} is:

Q = m \cdot L_\text{f}.

Apply this equation to find the size of the second part of energy input:

\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}.

Find the sum of these three parts of energy:

\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}.

You might be interested in
5. Which pair of words, in order, correctly completes
sattari [20]

Answer:

B is an answer.

3 0
3 years ago
A bus accelerates from 5.75 m/s at a rate of 1.25 m/s/s for 3.50
marysya [2.9K]

Answer:

10.125 meters?

Explanation:

Im taking 5.75m/s + 1.25 m/s/s (3.5) = my answer.

In those 3.5 seconds it travels 4.375.

I added that to 5.75 to get 10.125m

3 0
3 years ago
. An unbalanced force of 500 N is applied to a 75 kg object. What is the acceleration of the object?
Hoochie [10]

The acceleration of the object is 6.7 m/s^2

Explanation:

We can solve the problem by using Newton's second law, which states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:

F=ma

where

F is the net force

m is the mass of the object

a is its acceleration

For the object in this problem,

F = 500 N is the applied force

m = 75 kg is the force

Solving the equation for a, we find the acceleration:

a=\frac{F}{m}=\frac{500}{75}=6.7 m/s^2

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0
3 years ago
3. A microwave oven draws 12 A of current on a 110 V household circuit. What is its power
steposvetlana [31]

Answer:

W = 1320Watts

Explanation:

W = I*V

W = 12A*110V

W = 1320Watts

7 0
3 years ago
I'm willing to give a lot of points if u can help me out.
Anettt [7]

Answer:

i have absolutly no idea how to do it but i looked it up and your answer should be B. i could be wrong but thats what the web told me

7 0
3 years ago
Other questions:
  • Can been seen but not projected?
    6·1 answer
  • What’s the velocity of an 11-kilogram object with 792 joules of kinetic energy?
    12·2 answers
  • Ricardo is on vacation, doing some mountain climbing. He notices that the higher he goes up a mountain, the colder he feels. He
    13·2 answers
  • Waves will have the highest speed in_
    8·1 answer
  • Joe loves joe so yes​
    12·2 answers
  • What is a convex lens?​
    5·2 answers
  • A person pushes a 50kg box to the right with 100N of force at a constant velocity of 5m/s. Calculate the coefficient of friction
    14·1 answer
  • How many electron flow through a light bulb each second if the current flow through the light bulb 0.75A.The electric charge of
    12·1 answer
  • Arrange the objects in order from greatst to least of potential energy assume that gravity is constant
    15·1 answer
  • Calculate the speed of a proton after it accelerates from rest through a potential difference of 350 V.
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!