# Question b0ecf

Jul 12, 2017

Heat isn't required to cool water; water releases heat as it cools to its surroundings. Heat is energy.

#### Explanation:

I have no idea why they're giving us dimension of the cylinder? Maybe you're supposed to calculate the heat capacity of water that way, but that's silly relative to memorizing it, unless this is an important concept in your chemistry class.

First, we're looking to relate the heat capacity, temperature change, and mass of water required to perform the thermochemical change asked. Note: units are very important in these problems.

The equation for this is thus,

$q = m {C}_{s} \Delta T$

The variables you are given:
$m = 2000 k g \left(\frac{{10}^{3} g}{k g}\right) = 2.000 \cdot {10}^{6} g$
DeltaT = T_f-Ti = 8°C - 34°C = -26°C

The heat capacity of water is:
C_s = (4.184J)/(g*°C)

Relating these variables to each other:
q = (2.000*10^6g)* (4.184J)/(g*°C)*-26°C#
$q = - 2.2 \cdot {10}^{5} k J$

This is the heat transferred to the surroundings.