Answer
m = 10 kg = 10000g
$C_{s}$ = 4.18 $\frac{J}{g.K}$
$\Delta$$T$ = $T_{final}$ - $T_{initial}$ = 46.2$^{\circ}$C - 24.6$^{\circ}$C = 21.6$^{\circ}$C = 21.6 K
$q$ = ? $kJ$
$q$ = $C_{s}\times m \times \Delta T$
$q$ = $4.18 \frac{J}{g.K}\times 10000g \times 21.6 K $
$q$ = $902880 J $
$q$ = $902.88 kJ $
Therefore, in order to raise the temperature of 10.00 kg of liquid water from 24.6 to 46.2 °C, we need $902.88 kJ $
Work Step by Step
To find q, first we need the mass of water in g. Since the question gives us kg, we need to convert this value to g by multiplying it with 1000. The second thing that we need is the specific capacity of water which can be found in table 5.2. The last thing needed is the change in temperature in K which can be calculated by subtracting the initial temperature from the final temperature. When working with temperature changes, $^{\circ}$C and K are interchangeable. However, if we were to work with just temperature alone, there would be a conversion.
After finding all the necessary parts, we can find q by using this formula
$q$ = $C_{s}\times m \times \Delta T$. The last thing to do is converting our value in J to kJ since the question asks for that specific unit. This can be done by dividing the value in J by 1000.
