A transfer of heat within a liquid or gas that involves warm particles moving in currents is

Calculate the amount of heat needed to raise the temperature of 200g of ice from-30°C 50C water. (C = .5 for ice, C 1 for water,

Nitella [24]

Answer: The amount of heat needed is 29000 Cal.

Explanation:

The process involved in this problem are:

$(1):H_2O(s)(-30^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(50^oC)$

Now, we calculate the amount of heat released or absorbed in all the processes.

For process 1:

$q_1=mC_{p,s}\times (T_2-T_1)$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of ice = 200 g

$T_2$ = final temperature = $0^oC$

$T_1$ = initial temperature = $-30^oC$

Putting all the values in above equation, we get:

$q_1=200g\times 0.5Cal/g^oC\times (0-(-30))^oC=3000Cal$

For process 2:

$q_2=m\times L_f$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 200 g

$L_f$ = latent heat of fusion = 80 Cal/g

Putting all the values in above equation, we get:

$q_2=200g\times 80Cal/g=16000Cal$

For process 3:

$q_3=m\times C_{p,l}\times (T_{2}-T_{1})$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of water = 200 g

$T_2$ = final temperature = $50^oC$

$T_1$ = initial temperature = $0^oC$

Putting all the values in above equation, we get:

$q_3=200g\times 1Cal/g^oC\times (50-0)^oC=10000Cal$

Calculating the total heat absorbed, we get:

$Q=q_1+q_2+q_3$

$Q=3000+16000+10000=29000Cal$

Hence, the amount of heat needed is 29000 Cal.

7 0

3 years ago

Describe a situation when you might travel a high velocity but with low acceleration

Umnica [9.8K]

Answer:
One situation could be when someone is sitting in a car that is travelling at a constant speed of 150mph

Explanation:
Let’s review the terms:
Velocity: the speed of something in a given direction
Acceleration: increase in the rate or speed of something

The car is travelling at a high speed but the speed is constant, so it isn’t increasing. In other words, the car the person is sitting in is travelling in in a high velocity but with no/low acceleration.

I hope this helps! Please comment if you have any questions.

4 0

3 years ago

You observed three different star clusters and found that the main-sequence turnoff stars in cluster 1 had spectral type A, the

9966 [12]

Answer: Cluster 3 (D)

Explanation: Cluster 3 is the oldest and it could be the source from which cluster 2 and 3 got their spectral type from.

4 0

4 years ago

Read 2 more answers

Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 53 km/h. After six hour

Serhud [2]

For the answer to the question above, first find out the gradient.

m = rise/run 
=(y2-y1)/(x2-x1) 

the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" 
(x1,y1) = (3,53) 
(x2,y2) = (6,62) 

sub values back into the equation 
m = (62-53)/(6-3) 
m = 9/3 
m = 3 

now we use a point-slope form to find the the standard form 
y-y1 = m(x-x1) 
where x1 and y1 are any set of point given 
y-53 = 3(x-3) 
y-53 = 3x - 9 
y = 3x - 9 + 53 
y = 3x + 44 

y is the velocity of the car, x is the time.
I hope this helps.

4 0

3 years ago

Which of these characteristics would best help a student who is focused on reaching academic goals?

Anarel [89]

Answer: Responsibility

5 0

3 years ago