Calculating Temperature & Heat Transfer with a Piece of Metal

[always_smile]

A piece of metal with a mass of 1.50 kilograms, specific heat of 200 J/kg · C°, and initial temperature of 100° C is dropped into an insulated jar that contains liquid with a mass of 3.00 kilograms, specific heat of l,000 J/kg · C°, and initial temperature of 0° C. The piece of metal is removed after 5 seconds, at which time its temperature is 20° C. Neglect any effects of heat transfer to the air or to the insulated jar.

What is the temperature of the liquid after the metal is removed ?
a.) 0° C
b.) 4° C
c.) 8° C
d.) 10° C
e.) 20° C

What is the average rate at which heat is transferred while the piece of metal is in the liquid ?
a.) 4,000 J/s
b.) 4,800 J/s
c.) 6,000 J/s
d.) 9,600 J/s
e.) 16,000 J/s

i have no idea what i should do
please help me
thanks

 

[G01]

Read the forum rules please. We cannot help you unless you show some proof that you have put effort into this problem, i.e. work. You do not necessarily need to have equations and numbers, but you must show that you have put some time into the problem posted.

Try to answer the following and then we'll be able to help you here:

1) What concepts apply here?

2)What actually happened to cause the metal to cool down and the water to heat up?

3) Now that you have answered the above and have a list of concepts that apply, do you know of any formulas or equations that describe some of these concepts quantitatively. If so, list them.If you can give an attempt at answering some of these questions, then we'll be able to help you.

 

[ogb p]

I would approach this problem by using the equation Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.

To find the temperature of the liquid after the metal is removed, we can use the equation Q = mcΔT for both the metal and the liquid. Since the metal's initial temperature is 100°C and its final temperature is 20°C, we can calculate the heat transferred from the metal as follows:

Qmetal = (1.50 kg)(200 J/kg · C°)(20°C - 100°C) = -24,000 J

Since the heat lost by the metal is equal to the heat gained by the liquid, we can set Qmetal = Qliquid and solve for the final temperature of the liquid:

-24,000 J = (3.00 kg)(1000 J/kg · C°)(Tliquid - 0°C)
-24,000 J = 3,000,000 J/kg · C°(Tliquid)
Tliquid = -0.008°C

Since the temperature of the liquid cannot be negative, we can conclude that the final temperature of the liquid is 0°C. Therefore, the correct answer is option a) 0°C.

To find the average rate at which heat is transferred while the metal is in the liquid, we can use the equation Q/t = mcΔT/t, where t is the time. Since the metal was in the liquid for 5 seconds, we can calculate the average rate of heat transfer as follows:

Q/t = (1.50 kg)(200 J/kg · C°)(20°C - 100°C)/5 s = -4,800 J/s

Therefore, the correct answer is option b) 4,800 J/s.

In conclusion, as a scientist, I would use the appropriate equations and calculations to determine the temperature of the liquid after the metal is removed and the average rate of heat transfer during the process. It is important to carefully consider all the given information and use the correct units in the calculations to ensure accurate results.