The energy given off by 300 grams of an alloy as it cools through 50°C raises the temperature of 300 grams of water from 30°C to 40°C. The specific heat of the alloy (in cal/g C ) is:

i know Q=CpM(delta T)

but what is what?

Responses

* physics - bobpursley, Monday, June 22, 2009 at 8:22pm

Heat lost=heat gained
300*c*50=300*4.18*10

I did this, but i got .836, and the answer is suppose to be .2 so im not sure what i'm doing wrong!!

Do in Cals, not Joules so 300*c*50= 300*1*10, which will give you 3000/15000= .2

This is a simple math problem .

c=300/300 * 4.18* 10/50 = .8 j/gC about.

Now if you do specific heat in calories, you get .2. Numbers without units are meaningless.

Well, it seems like you're feeling a bit heated about this problem! Let me break it down for you with a dash of humor.

First, let's identify what we know. The heat lost by the alloy is equal to the heat gained by the water. This gives us the equation:

Q(lost) = Q(gained)

Now, we can plug in the values we have. The mass of the alloy and water are both 300 grams. The change in temperature for the alloy is 50°C (cooling from 50°C to 0°C) and for the water, it is 10°C (heating from 30°C to 40°C).

So, our equation becomes:

C(alloy) * M(alloy) * ΔT(alloy) = C(water) * M(water) * ΔT(water)

Now, let's substitute in the known values:

C(alloy) * 300g * 50°C = 4.18 cal/g°C * 300g * 10°C

And let the clownery begin!

C(alloy) * 15000 = 12540

Now, divide both sides by 15000:

C(alloy) = 12540 / 15000

And the laughter begins to fade as we calculate:

C(alloy) = 0.836

Oh no, it seems like our answer doesn't match the expected value of 0.2! You might want to double-check your calculations, or if this is a multiple-choice question, consider if there could be a typo in the possible answers. Those sneaky typos can really throw us off, can't they? Keep trying, and don't let this hiccup dampen your spirits. Remember, laughter is the best seasoning for any problem!

Let's go step by step to solve the problem.

Given:
Mass of the alloy (M1) = 300 grams
Change in temperature of the alloy (delta T1) = -50°C (since it is cooling)
Mass of water (M2) = 300 grams
Change in temperature of water (delta T2) = 10°C (from 30°C to 40°C)
Specific heat capacity of water (C2) = 4.18 cal/g°C

We need to find the specific heat capacity of the alloy (C1).

The heat gained by the water is equal to the heat lost by the alloy, so we can write the equation as:

heat lost by the alloy = heat gained by the water

Using the formula Q = Cp * M * delta T, we can write the equation as:

C1 * M1 * delta T1 = C2 * M2 * delta T2

Substituting the values we have:

C1 * 300 * (-50) = 4.18 * 300 * 10

Now, let's solve for C1:

C1 * (-15000) = 12540

C1 = 12540 / (-15000)

C1 ≈ -0.836 cal/g°C

It seems you made a sign mistake while plugging in the values. The specific heat capacity of the alloy is approximately -0.836 cal/g°C, not 0.836 cal/g°C.

So, it appears that there might be an error in the answer you were given. The specific heat capacity of the alloy cannot be negative, as it represents the amount of energy required to raise the temperature of the alloy by 1 degree Celsius per gram.

To find the specific heat of the alloy, you can use the equation Q = Cp * m * ΔT, where Q is the amount of heat transferred, Cp is the specific heat capacity of the material, m is the mass of the material, and ΔT is the change in temperature.

In this case, the heat lost by the alloy is equal to the heat gained by the water. The mass of the alloy and water are both given as 300 grams. The change in temperature for the alloy is 50°C, and for the water is 10°C (from 30°C to 40°C).

So, the equation becomes: Cp * 300 * 50 = 4.18 * 300 * 10

To solve for Cp, divide both sides of the equation by (300 * 50):
Cp = (4.18 * 300 * 10) / (300 * 50)

Simplifying the expression further:
Cp = (4.18 * 10) / 50
Cp = 0.836 cal/g°C

It seems that you made a calculation error in your previous attempt. The correct answer should indeed be 0.836 cal/g°C, not 0.2 cal/g°C.