Question

Question #fcc68

Answer 1

About `"5 cal"`, or about `"20 J"`. (The more exact value was `"20.92 J"`, but the `"5 cal"` is not rounded.)

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The heat energy `q` required to raise the temperature of `"1 g"` of water by `1^@ "C"` is one calorie, `"1 cal"`. That is the definition of the specific heat capacity `c` for water, `c_w`:

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

And thus, to raise the temperature of a gram of water by `5^@ "C"`, we assume the specific heat capacity of water stays constant in this temperature range and write

`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.

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And in case you want this in different units, note that in terms of joules, `"J"`, there are `"4.184 J"` in one `"cal"` at `25^@ "C"`, so we get approximately:

`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 `color(blue)("20 J")`. We could denote the digits past the last significant one as `20._(92)`.

(It would appear that we round down in the tens place for the wrong reason, but since `0` in the ones place (the digit past our last significant digit) is less than `5`, we do round down.)

Answer 2

`"20 J"`

Explanation:

The key here is the amount of heat needed to increase the temperature of `"1 g"` of water by `1^@"C"` `->` the 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 `"1 g"` of that substance by `1^@"C"`.

In the case of water, you need `"4.18 J"` of heat to increase the temperature of a single gram of water by `1^@"C"`.

This implies that in order to increase the temperature of `"1 g"` of water by `5^@"C"`, you need `5` times as much heat as you'd need to increase the temperature of `"1 g"` of water by `1^@"C"`.

`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.