Question

How many kilojoules of energy are needed to raised the temp of 400 grams of water from -10 C° to 114 C°? (5 steps)

Answer 1

Answer:

207.33 kJ.

Explanation:

What is given?

Mass of water (m) = 400 g.

ΔT (Change of temperature) = 114 °C - ( - 10 °C) = 124 °C.

Specific heat of water (c) = 4.18 J/g°C.

Step-by-step solution:

To solve this problem we have to use the following formula:

$Q=m\cdot c\cdot\Delta T.$

Where Q is heat, m is the mass, c is the specific heat of the water and ΔT is the change of temperature ( T final - T initial).

We want to find the value of Q, so we replace the data that we have:

$\begin{gathered} Q=400\text{ g}\cdot418\frac{J}{g\text{ \degree C}}\cdot124\text{ \degree C,} \\ Q=207328\text{ J.} \end{gathered}$

But the problem is asking the answer in kilojoules, so remember that 1 kJ equals 1000 J, so the conversion will look like this:

$207328\text{ J}\cdot\frac{1\text{ kJ}}{1000\text{ J}}=207.33\text{ kJ.}$

The answer would be 207.33 kJ.