# Specific Heat Capacity

### By Connor Pike

## What is the Specific Heat Capacity of a Substance?

## What are the Units for Each Part of the Equation?

**°**C); and △T, or the change in temperature, is measured in degrees Celsius (

**°**C).

## Example Problems

**°**C to 93

**°**C? (the specific heat of water is 4.18J/g

**°**C)

Step 1: You are looking for the q value, or the joules of heat, so rearrange the equation to where q is left alone. After rearranging, your equation should be q=mc△T.

Step 2: Insert the values of each part of the equation except for q. Your equation should end up being q=(500g)(4.18J/g**°**C)(58**°**C).

Step 3: Multiply the values. You should end up with q=121,220J.

Step 4: Check for significant figures. Since 500 has only one sig fig, you would need to round 121,220 to 100,000. Your final answer will be amount of heat=100,000 joules.

Example Problem 2. What is the specific heat capacity of a substance that absorbs 7537 joules of heat when a sample of 223 grams of the substance increases in temperature by 50**°**C?

Step 1: You are looking for the c value, or the specific heat capacity, so rearrange the equation to where c is left alone. After rearranging, your equation should be c=q/m△T.

Step 2: Insert the values of each part of the equation except for c. Your equation should end up being c=(7537J)/(223g)(50.0**°**C).

Step 3: Multiply the values. You should end up with c=0.67596412556J/g**°**C.

Step 4: Check for significant figures. Since 50.0 and 223 have only three sig figs, you would need to round 0.67596412556 to 0.676. Your final answer will be specific heat capacity=0.676J/g**°**C.