# The temperature of 50.0 g of water was raised to 50.0°C by the addition of 1.0 kJ of heat energy. What was the initial temperature of the water?

Nov 25, 2015

${45.2}^{\circ} \text{C}$

#### Explanation:

The idea here is that you need to use the mass of water, the heat added to the sample, and the specific heat of water to find the resulting change in temperature.

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

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

$q$ - heat absorbed
$m$ - the mass of the sample
$c$ - the specific heat of water, equal to 4.18"J"/("g" ""^@"C")
$\Delta T$ - the change in temperature, defined as final temperature minus initial temperature

Plug in your values and solve for $\Delta T$ - do not forget to convert the added heat from kilojoules to Joules

$q = m \cdot c \cdot \Delta T \implies \Delta T = \frac{q}{m \cdot c}$

DeltaT = (1.0 * 10^(3)color(red)(cancel(color(black)("J"))))/(50.0color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ""^@"C")) = 4.785^@"C"

So, if the temperature of the water changed by ${4.785}^{\circ} \text{C}$, and its final temperature is now ${50.0}^{\circ} \text{C}$, it follows that its initial temperature was

$\Delta T = {T}_{\text{f" - T_"i" implies T_"i" = T_"f}} - \Delta T$

T_"i" = 50.0^@"C" - 4.785^@"C" = color(green)(45.2^@"C")