If 200 grams of water is to be heated from 24°C to 100°C to make a cup of tea, how much heat must be added?

The specific heat of water is 4.18 $J$$/$$g \cdot C$.

Dec 23, 2015

$q = \text{64 kJ}$

Explanation:

As you know, a substance's specific heat tells you how much heat is required to increase the temperature of $\text{1 g}$ of that sample by ${1}^{\circ} \text{C}$.

In water's case, you know that its specific heat is equal to 4.18"J"/("g" ""^@"C"). This tells you that in order to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$, you need to supply $\text{4.18 J}$ of heat.

How much heat would you need to increase the temperature of $\text{200 g}$ of water by ${1}^{\circ} \text{C}$?

Well, if you need $\text{4.18 J}$ per gram to increase its temperature by ${1}^{\circ} \text{C}$, it follows that you will need $200$ times more heat to get this done.

Likewise, if you were to increase the temperature of $\text{1 g}$ of water by ${76}^{\circ} \text{C}$, you'd need $76$ times more heat than when increasing the temperature of $\text{1 g}$ by ${1}^{\circ} \text{C}$.

Combine these two requirements and you get the total amount of heat required to increase the temperature of $\text{200 g}$ of water by ${76}^{\circ} \text{C}$.

Mathematically, this is expressed using the following equation

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

$q$ - heat absorbed/lost
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as final temperature minus initial temperature

Plug in your values to get

$q = 200 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (100 - 24)color(red)(cancel(color(black)(""^@"C}}}}$

$q = \text{63536 J}$

I'll leave the answer rounded to two sig figs and expressed in kilojoules

$q = \textcolor{g r e e n}{\text{64 kJ}}$