Lesson 1.4 - Specific Heat Capacity Calculations

Now, you can try some that are a little more difficult.

Example 1:

According to the table, the specific heat capacity of motor oil is 2.00 J/g" °C.  How much heat would 4.00 g of motor oil absorb if the temperature was increased by 1.00°C?

Solution:

You will use the same formula as on the previous page:

q = mcDt

in which   

q is the heat  (the unknown in this example)
m is the mass (4.00 g)
c is the specific heat capacity (2.00 J/g" °C)
Dt  is the change in temperature (1.00°C)

Now, enter all the numbers into the formula:

q = mcDt  
q  = (4.00 g)´(2.00 J/g" °C)´(1.00°C)
q  = 8.00 J

Did you see how the calculation was done?  The three numbers were multiplied.  Try it yourself.  Then, try this one.

Question 1. How much heat would 3.00 g of vegetable oil absorb if the temperature was increased by 1.00°C?

Then, return here to continue this lesson.

 

 

Did you get it right?  Congratulations if you did.  If you got it wrong, go over the calculation again to be sure you understand how it is done before you try the following two questions.  You might even have to phone for help. 

Question 2. How much heat would 5.00 g of glass absorb if the temperature was increased by 1.00°C?

Question 3. How much heat would 15.00 g of sand absorb if the temperature was increased by 1.00°C? 

 

Then, return here to continue this lesson.

 

 

 

 

 

 

 

 

 

 

 

 

 

Answers to Questions:

Question 1. How much heat would 3.00 g of vegetable oil absorb if the temperature was increased by 1.00°C?

q = mcDt 
q = (3.00 g)´(1.97 J/g" °C)´(1.00°C)
q = 5.91 J

 

Click to return to where you left off in this lesson.

 

 

 

 

 

 

 

 

 

 

 

Answers to Questions:

Question 2. How much heat would 5.00 g of glass absorb if the temperature was increased by 1.00°C?

q = mcDt 
q = (5.00 g)´(0.84 J/g" °C)´(1.00°C)
q = 4.2 J

Question 3. How much heat would 15.00 g of sand absorb if the temperature was increased by 1.00°C? 

q = mcDt 
q = (15.00 g)´(0.66 J/g" °C)´(1.00°C)
q = 9.9 J

 


Go to the next page to continue Lesson 1.