Specific heat of

Question

A 25.00 gram sample of an unknown metal initially at 99.0 degrees Celcius is added to 50.00 grams of water initially at 10.7 degrees Celcius. The final temperature of the system is 20.15 degrees Celcius. Calculate the specific heat of the metal. 

Answer 1

To solve this question, we will use the equation Q = m*c*ΔT. Remember that Q is the amount of heat that is required to change the temperature. m is the mass of the substance in kilograms (kg). c is the specific heat capacity of the substance. Remember that the specific heat capacity of water is 4,184 J/(kg*C). If you don't already know this, you should memorize it, because problems such as this one depend on you knowing the specific heat capacity for water. ΔT means the change in temperature.

Remember that ΔT = final temperature -- initial temperature.

In this problem we have two substances: the unknown compound and the water. Remember that the final temperature is the same for both the unknown compound and the water. Also, remember that the amount of heat lost by the unknown compound (which loses heat because it cools down from 99.0 Celsius to the final temperature of 20.15 degrees Celsius) is the same as the amount of heat gained by the water (which gains heat because it heats up from 10.7 degrees Celsius to 20.15 degrees Celsius.

In other words, Q of the unknown compound = Q of the water. To find the Q of the unknown compound, we should find the Q of the water. Let's review the variables we know for the water:

c = 4,182 J/(kg*C)

m = 50.00 g; since 1000 g = 1 kg, m = 0.05 kg

initial température of the water = 10.7 degrees Celsius

final temperature of the water = 20.15 degrees Celsius

ΔT = final temperature -- initial temperature = 20.15 -- 10.7, so ΔT = 9.45

Q = what we want to find

Now we can put the information into the equation Q = m*c*ΔT like this:

Q = 0.05 * 4,182 * 9.45

We just multiply the numbers. The calculator says 0.05 * 4,182 * 9.45 = 1975.995.

So with the proper unit of Joules, Q for the water = 1975.995 J. This is the Q for the water. The Q for water is positive since the water gains heat. For the unknown metal, Q is negative because the metal is losing heat. So the Q for the unknown metal = -1975.995 J. Now we can use this value to find the specific heat capacity (c) of the unknown metal. Let's review the values we know for the unknown metal:

Q = -1975.995 J

m = 25.00 g; since 1000 g = 1 kg, m = 0.025 kg

c = what we want to find

initial temperature of the unknown metal = 99.0 degrees Celsius

final temperature of the unknown metal = 20.15 degrees Celsius

ΔT = 20.15 -- 99.0 = -78.85; so ΔT = -78.85

We put these values into the equation Q = m*c*ΔT like this:

-1975.995 = 0.025 * c * -78.85

Let's first multiply 0.025 * -78.85. The calculator says that 0.025 * -78.85 = -1.97125 . So we replace 0.025*-78.85 with -1.97125 like this:

-1975.995 = -1.97125 * c

Now we find c by dividing both sides by -1.97125. Our equation looks like this:

c = -1975.995 / -1.97125

We use a calculator to calculate -1975.995 / -1.97125 = 1002.4.

With the proper unit, c = 1002.4 J/(kg*C).

So the specific heat capacity of the unknown metal is 1002.4 J/(kg*C)