# Question 38a0a

Jun 6, 2015

The man released 1.8 MJ of heat into the water.

The equation that establishes a relationship between the heat absorbed by the water and the change in temperature looks like this

$q = m \cdot c \cdot \Delta T$, where

$q$ - the heat supplied by the man to the water;
$m$ - the mass of the water;
$c$ - the specific heat of water, equal to $\text{4.18 J/g"^@"C}$;
$\Delta T$ - the change in temperature, defined as the diffference between the final temperature and the initial temperature of the water.

So, the first thing you need to do is determine the mass of water present in the tub. Since no mention of water's density was made, you can assume it to be equal to $\text{1 g/mL}$.

This means that you have

12750cancel("mL") * "1 g"/(1cancel("mL")) = "12750 g water"#

A substance's specific heat tells you how much energy you need to supply to 1 gram of that substance in order to increase its temperature by ${1}^{\circ} \text{C}$.

In water's case, you need 4.18 J to increase the temperature of 1 gram by ${1}^{\circ} \text{C}$. Since you need to raise the temperature of a significantly bigger mass of water by much more than ${1}^{\circ} \text{C}$, you're going to need a lot of heat.

Plug your values into the equation and solve for $q$

$q = 12750 \cancel{\text{g") * 4.18"J"/(cancel("g")^@cancel("C")) * (86-52)^@cancel("C}}$

$q = \text{1,812,030 J}$

Rounded to two sig figs, the number of sig figs you gave for the two temperatures of the water, and aexpressed in megajoules, the answer will be

$1 , 812 , 030 \cancel{\text{J") * "1 MJ"/(10^(6)cancel("J")) = color(green)("1.8 MJ}}$