What is the final temperature of the soup after the ice has melted?

[mustang]

Problem 9.
A student drops metallic objects into a
224 g steel container holding 325 g of water
at 28±C. One object is a 472 g cube of copper
that is initially at 87 degrees C, and the other is a
chunk of aluminum that is initially at 9 degreesC.
To the students's surprise, the water reaches
a final temperature of 28 degreesC, precisely where
it started.
What is the mass of the aluminum chunk?
Assume the specifc heat of copper and alu-
minum are 387 J/kg degrees C
± C and 899 J/kg degrees C

Note: Would you used: m*c*delta t = m*c*delta t
If so what's next?

Problem 12.
Given: specific heat of water = 4186 J/kg degrees Ceicuis
and water's latent heat of fusion = 3.33 *10^5 J/kg.

A 0.012 kg cube of ice at 0.0degrees Ceicuis is added to
0.459 kg of soup at 80.4degrees Ceicuis.
Assuming that the soup has the same specific
heat capacity as water, find the final tem-
perature of the soup after the ice has melted.
Answer in units of degrees Ceicuis.

 

[Doc Al]

Problem 9.
...
Note: Would you used: m*c*delta t = m*c*delta t
If so what's next?

Set it up and solve!
[tex]m_1c_1\Delta t_1 = m_2c_2\Delta t_2[/tex]

Problem 12.
...
Assuming that the soup has the same specific
heat capacity as water, find the final tem-
perature of the soup after the ice has melted.

Realize that whatever heat is removed from the soup, goes into (1) melting the ice (2) heating up the water from that ice.

 

[mustang]

some help!

Problem 011. A sample of lead used to make a lead sinker
for fshing has an initial temperature of 25.7degrees C
and is poured into a mold immediately after
it has melted.
How much energy is needed to melt 0.274
kg of lead? Assume the specifc heat, the
latent heat and the melting point of lead are
128 J/kg * degrees C, 2.45*10^4 J/kg and 327.3 degreesC
respectively. Answer in units of J.

Note: Would it be done this way
1. delta T=100-25.7=74.3
Q1=(0.274kg)(128)=74.3
Is this right? If not what should I do?

Problem 13.When a driver brakes an automobile, fric-
tion between the brake disks and the brake
pads converts part of the car's translational
kinetic energy to internal energy.
If a 1610 kg automobile traveling at 28 m/s
comes to a halt after its brakes are applied,
how much can the temperature rise in each
of the four 3.8 kg steel brake disks? Assume
the disks are made of iron (cp = 448 J/kg *
degrees C)
and that all of the kinetic energy is distributed
in equal parts to the internal energy of the
brakes. Answer in units of degrees C.

 

[Doc Al]

Problem 011. A sample of lead used to make a lead sinker
for fshing has an initial temperature of 25.7degrees C
and is poured into a mold immediately after
it has melted.
...
Note: Would it be done this way
1. delta T=100-25.7=74.3
Q1=(0.274kg)(128)=74.3
Is this right? If not what should I do?

Do it step by step. First find the energy needed to raise the temperature to the melting point. (Note: the melting point is not 100°!) Then find the energy needed to melt the lead.

Problem 13.When a driver brakes an automobile, fric-
tion between the brake disks and the brake
pads converts part of the car's translational
kinetic energy to internal energy.
...

Start by finding how much heat is generated: Find the KE of the car.

 

[mustang]

rds to problem 12

I used
m*c*delta T=m*c*delta T

where I used the c as 4186J/kg*C for both of the specific heats. Is this right?

 

[mustang]

I have solved all of my problems for my homework except for 12 and 13. Where I am clueless. Need step-by-step help.

-Logic is a systematic method of coming to the wrong conclusion with confidence.
(Manley's Maxim)