# Question #fcc68

##### 2 Answers

About

The *heat energy* **specific heat capacity**

#c_w = "1 cal/g"^@ "C"#

And thus, to raise the temperature of a gram of water by

#color(blue)(q) = ("1 cal")/(cancel"g"cancel(""^@ "C")) xx 5cancel(""^@ "C") xx cancel"1 g" ~~ color(blue)("5 cal")#

This is from an actual equation in disguise...

#q = m_wc_wDeltaT# where we assigned the mass of water as

#m_w = "1 g"# and the change in temperature as#DeltaT = T_2 - T_1# #=# #5^@ "C" - 0^@ "C"# #=# #5^@ "C"# .

But we could have figured it out from the units if the equation does not come to mind, as you saw above.

And in case you want this in different units, note that in terms of joules,

#5 cancel"cal" xx "4.184 J"/cancel"cal" ~~ ul"20.92 J"#

to one significant figure, as you have given, it would end up being

(It would appear that we round down in the tens place for the wrong reason, but since *past* our last significant digit) is less than

#### Answer:

#### Explanation:

The key here is the amount of heat needed to increase the temperature of **specific heat** of water.

#color(blue)(ul(color(black)(c_"water" = "4.18 J g"^(-1)""^@"C"^(-1))))#

So, the specific heat of a substance tells you the amount of heat needed in order to increase the temperature of

In the case of water, you need

This implies that in order to increase the temperature of **times as much heat** as you'd need to increase the temperature of

#5 color(red)(cancel(color(black)(""^@"C"))) * overbrace("4.18 J"/("1 g" * 1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("the specific heat of water")) = color(darkgreen)(ul(color(black)("20 J")))#

The answer must be rounded to one **significant figure**.