# The specific heat capacity of liquid water is 4.18 kJ/g C, how would you calculate the quantity of energy required to heat 1.00 g of water from 26.5 C to 83.7 C?

##### 1 Answer

#### Explanation:

First thing first, you mistyped the specific heat of water, which should be

#c_"water" = 4.18"J"/("g" ""^@"C")#

Now, a substance's *specific heat* tells you how much heat is required to increase the temperature of

In the case of water, you would need

Notice that your sample of water has a mass of

The equation that establishes a relationshop between heat and change in temperature looks like this

#color(blue)(q = m * c * DeltaT)" "# , where

*final temperature* and the *initial temperature*

Plug in your values and solve for

#q = 1.00color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * (83.7 - 26.5)color(red)(cancel(color(black)(""^@"C")))#

#q = "239.096 J"#

Rounded to three sig figs, the answer will be

#q = color(green)("239 J")#

**SIDE NOTE** *Notice how much heat would be needed to increase the temperature of 57.2 g of water by* *will now be equal to* *and you'd have*

#q = 57.2color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * 1color(red)(cancel(color(black)(""^@"C")))#

#q = "239 J"#

This tells you that you need the same amount of heat to increase the temperature of