How Much Energy and Cost Are Involved in Heating Daily Water to 75°C?

[miniradman]

Homework Statement

A family uses 150 L of hot water each day. The hot water system is set to 75°C and the
supply water has a temperature of 20°C. How much energy is required? If the hot water
consumption was spread out evenly over the whole day (24 hrs), what is the total daily cost
(assuming 14 c/kWh)?

Homework Equations

$Q=mC \Delta T$

$C = 4.184 kJ/kg (HeatingCapacity)$

The Attempt at a Solution

Hi all, I have this question and I thought I was doing it correctly, but for some reason I'm not getting the required answer.
$density of water = 998$

$Volume = 0.15 m^{3}$

$m = 149.7 kg$

$\Delta T = 55$

$Q=(149.7)(4.184)(55)$
$Q=34424.26 kJ$

$seconds/day = 86400s$

$p = \frac{dQ}{dT} = \frac{\Delta Q}{\Delta T}$

$p = \frac{34424.26 kJ }{86400s}$

$p = 0.39843kW$

$p kWh = 0.39843kW (3600)$

$p kWh = 1434kWh$

$cost = (0.14c) (1434kWh)$

$cost = 200c = 2.00$

However the answer states that it is $1.35/day

I checked my work three times through, but I think I have a fundamental misunderstanding of something which is stopping me from getting the correct answer.

 

[DrClaude]

$p = 0.39843kW$

$p kWh = 0.39843kW (3600)$

$p kWh = 1434kWh$

This is where you go wrong. In that second line, you are actually calculating how much energy is used every hour:
$$
0.39843\ \textrm{kW} \times \frac{3600\ \textrm{s}}{\textrm h} = 1434\ \textrm{kJ/h}
$$
Instead, find the conversion factor to go from kJ to kWh:
$$
\begin{align}
1 \textrm{kWh} &= 1 \textrm{kW} \times \textrm{h} \\
&= 1 \frac{\textrm{kJ}}{ \textrm{s}} \times \textrm{h} \\
&= 1 \frac{\textrm{kJ}}{ \textrm{s}} \times \textrm{h} \times \frac{3600\ \textrm{s}}{\textrm h} \\
&= 3600\ \textrm{kJ}
\end{align}
$$

 

[miniradman]

Ahh, I plugged that conversion factor in after I found Q, then got $1.33... close enough?

Cheers for the response