A piece of unknown metal w/ a mass of 23.8 g is heated to 100.0 degrees celcius and dropped into 50.0 ml of...

Question

A piece of unknown metal w/ a mass of 23.8 g is heated to 100.0 degrees celcius and dropped into 50.0 ml of water at 24.0 degrees celcius. The final temperature of the system is 32.5 degrees celcius. What is the specific heat of the metal? (Hint : the energy lost by one system is equal to the energy gained by the other system). Please show all your work.

Answer 1

Ok, this is my first time answering a chem. question, but I think I got this one.

The formula to calculate energy is q= (m)(cp)(delta T)
q is the energy in Joules or calories. m is mass in grams. cp is the specific heat in J/g degrees Celcius. delta t is change in temp.

The formula for delta T is change in temp. = final temp. subtract initial temp.

So you figure out the water first because it's specific heat is 4.184 J/g degrees Celcius and the temperature change is 32.5 - 24.0, so 8.50 degree Cel.

For the mass, you have to convert the ml to grams, so mass = density times volume. Density of H2O is 1.00 g/ml so multiplied with 50.0 ml is 50.0 grams. So you input the mass, specific heat, and delta temp. values into the energy equation to get the energy of the water.

Now the hint says, the energy lost from one system is gained by the other system. So that means that whatever positive energy value you got as an answer for the water is going to be negative energy value for the metal. Because energy gained is a positive q value and energy lost is a negative q value.

So the equation to calculate specific heat is : q/ (m)(delta T)

so cp= -q/ 23.8 (32.5-100.0) (notice how the q value is negative in this case because the metal lost heat.)

Your answer should be in the unit J/g degrees Celcius.

Answer 2

I think the equation you have is wrong. To find the amount of joules gained or lost, it should be (specific heat)/(mass)(temperature). Also, I believe that in the second part of the problem to find the specific heat of the unknown substance, it should be divided by 92 degrees Celsius instead of 72. This is because the substance did not lose 28 degrees, it only lost 8.

Answer 3

The formula is q = m Cp delta T

q = heat, m = mass, Cp = specific heat and delta T is final T - Initial T

In your problem start with the water. It will gain heat. 50 ml of water = 50 grams.

q = (50g) (1) (32.5-24.0)

Now take this q as a minus (gives up heat) and use the same equation to solve for the Cp of the metal.

m = 23.8 g and delta T = (32.5-100.0)

Solve for Cp and you are done. The units will be cal/ gr oC

Answer 4

well...

First, you check the change in thermal energy of the water:

E = mc(change in t)

where, m equals mass, and c equals specific heat:

E= (.050 kg)(4.18)(32.5-24)
E= 17.8 kJ

Whatever energy water gains is the amount of energy that the metal loses. So you use the same equation:

E=mc(change in t)
-17.8 = (.0238 kg) (c) (32.5-100)
-17.8 = -1.6065c
11.07 = c

So the specific heat of the metal is 11.07 kJ per Kelvin per kilogram, or 11,070 Joules per Kelvin per gram