# If 100.00 J are added to 20.0 g water at 30.0°C, what will be the final temperature of the water?

Jan 5, 2017

The final temperature of the water will be 31.2 °C.

#### Explanation:

The formula for the heat gained or lost by a substance is

color(blue)(bar(ul(|color(white)(a/a)q = mcΔTcolor(white)(a/a)|)))" "

We can rearrange this equation to get

ΔT = q/(mc)

In this problem,

$q = \text{100.00 J}$
$m = \text{20.0 g}$
$c = \text{4.178 J·K"^"-1""g"^"-1}$

ΔT = (100.00 color(red)(cancel(color(black)("J"))))/(20.0 color(red)(cancel(color(black)("g"))) × "4.178 J"·color(red)(cancel(color(black)("K"^"-1""g"^"-1")))) = "1.197 °C"

ΔT = T_"f" - T_"i" = T_"f"color(white)(l) "- 30.0 °C" = "1.197 °C"

${T}_{\text{f" = "30.0 °C + 1.197 °C" = "31.2 °C}}$