Below is a full, step-by-step estimate of the energy and cost to heat stationary water in a 15 mm copper pipe, 15 m long, from 10 °C to 55 °C, with ambient air at 10 °C, no insulation, electricity at $0.28/kWh.

I will include both the energy to heat the water and the copper pipe wall, since you specified copper.

1. Geometry and assumptions

1.1 Pipe dimensions

In plumbing, a “15 mm copper pipe” is normally a 15 mm outside diameter (OD) pipe.

We must assume a wall thickness to get the internal diameter. A common wall thickness is about 0.7 mm, so:

• Outside radius Ro=7.5 mm=0.0075 mR_o = 7.5 \text{ mm} = 0.0075 \text{ m}Ro​=7.5 mm=0.0075 m

• Wall thickness $t\approx 0.7\text{ mm}=0.0007\text{ m}$

• Inside radius Ri=Ro−t=7.5−0.7=6.8 mm=0.0068 mR_i = R_o – t = 7.5 – 0.7 = 6.8 \text{ mm} = 0.0068 \text{ m}Ri​=Ro​−t=7.5−0.7=6.8 mm=0.0068 m

• Pipe length $L=15\text{ m}$

2. Volume and mass of water in the pipe

2.1 Volume of water

The water fills the internal cylinder:

Vwater=πRi2LV_{\text{water}} = \pi R_i^2 LVwater​=πRi2​L Vwater=π×(0.0068)2×15≈0.00218 m3V_{\text{water}} = \pi \times (0.0068)^2 \times 15 \approx 0.00218 \text{ m}^3Vwater​=π×(0.0068)2×15≈0.00218 m3

So:

• Vwater≈0.00218 m3V_{\text{water}} \approx 0.00218 \text{ m}^3Vwater​≈0.00218 m3

• 1 m31 \text{ m}^31 m3 of water ≈ 1000 kg

• ⇒mwater≈0.00218×1000≈2.18 kg\Rightarrow m_{\text{water}} \approx 0.00218 \times 1000 \approx 2.18 \text{ kg}⇒mwater​≈0.00218×1000≈2.18 kg

So the pipe holds about 2.2 L of water.

3. Energy to heat the water

Given:

• Initial temperature T_1 = 10 \,^\circ\text{C}

• Final temperature T_2 = 55 \,^\circ\text{C}

• Temperature rise ΔT=T2−T1=45 K\Delta T = T_2 – T_1 = 45 \, \text{K}ΔT=T2​−T1​=45K

• Specific heat of water cwater≈4.186 kJ/kg\cdotpKc_{\text{water}} \approx 4.186 \text{ kJ/kg·K}cwater​≈4.186 kJ/kg\cdotpK

Energy to heat water:

Qwater=mwater cwater ΔTQ_{\text{water}} = m_{\text{water}} \, c_{\text{water}} \, \Delta TQwater​=mwater​cwater​ΔT Qwater≈2.18×4.186×45 kJQ_{\text{water}} \approx 2.18 \times 4.186 \times 45 \, \text{kJ}Qwater​≈2.18×4.186×45kJ Qwater≈410 kJ(more precisely ≈410.5 kJ)Q_{\text{water}} \approx 410 \, \text{kJ} \quad (\text{more precisely } \approx 410.5 \text{ kJ})Qwater​≈410kJ(more precisely ≈410.5 kJ)

Convert to kWh (since $1\text{ kWh}=3,600\text{ kJ}$):

Ewater=410.53600≈0.114 kWhE_{\text{water}} = \frac{410.5}{3600} \approx 0.114 \text{ kWh}Ewater​=3600410.5​≈0.114 kWh

So heating just the water costs, at 28 ¢/kWh:

Costwater=0.114×0.28≈0.032 dollars\text{Cost}_{\text{water}} = 0.114 \times 0.28 \approx 0.032 \text{ dollars}Costwater​=0.114×0.28≈0.032 dollars

≈ 3.2 cents.

4. Energy to heat the copper pipe

You explicitly said the pipe is copper and uninsulated, so we should also heat the pipe wall from 10 °C to 55 °C.

4.1 Volume and mass of copper

Copper volume is the volume of the annulus (outer cylinder minus inner cylinder):

VCu=π(Ro2−Ri2)LV_{\text{Cu}} = \pi (R_o^2 – R_i^2) LVCu​=π(Ro2​−Ri2​)L VCu=π[(0.0075)2−(0.0068)2]×15V_{\text{Cu}} = \pi \big[(0.0075)^2 – (0.0068)^2\big] \times 15VCu​=π[(0.0075)2−(0.0068)2]×15 VCu≈4.72×10−4 m3V_{\text{Cu}} \approx 4.72 \times 10^{-4} \text{ m}^3VCu​≈4.72×10−4 m3

Density of copper ρCu≈8960 kg/m3\rho_{\text{Cu}} \approx 8960 \text{ kg/m}^3ρCu​≈8960 kg/m3:

mCu=ρCuVCu≈8960×4.72×10−4≈4.23 kgm_{\text{Cu}} = \rho_{\text{Cu}} V_{\text{Cu}} \approx 8960 \times 4.72 \times 10^{-4} \approx 4.23 \text{ kg}mCu​=ρCu​VCu​≈8960×4.72×10−4≈4.23 kg

So there is about 4.2 kg of copper in that 15 m run.

4.2 Energy to heat the copper

• Specific heat of copper cCu≈0.385 kJ/kg\cdotpKc_{\text{Cu}} \approx 0.385 \text{ kJ/kg·K}cCu​≈0.385 kJ/kg\cdotpK

• Same temperature rise $\Delta T=45\text{ K}$

QCu=mCu cCu ΔTQ_{\text{Cu}} = m_{\text{Cu}} \, c_{\text{Cu}} \, \Delta TQCu​=mCu​cCu​ΔT QCu≈4.23×0.385×45 kJQ_{\text{Cu}} \approx 4.23 \times 0.385 \times 45 \, \text{kJ}QCu​≈4.23×0.385×45kJ QCu≈73.2 kJQ_{\text{Cu}} \approx 73.2 \text{ kJ}QCu​≈73.2 kJ

Convert to kWh:

ECu=73.23600≈0.0203 kWhE_{\text{Cu}} = \frac{73.2}{3600} \approx 0.0203 \text{ kWh}ECu​=360073.2​≈0.0203 kWh

5. Total energy and cost

5.1 Total energy to heat water + copper

Qtotal=Qwater+QCu≈410.5+73.2=483.7 kJQ_{\text{total}} = Q_{\text{water}} + Q_{\text{Cu}} \approx 410.5 + 73.2 = 483.7 \text{ kJ}Qtotal​=Qwater​+QCu​≈410.5+73.2=483.7 kJ Etotal=483.73600≈0.134 kWhE_{\text{total}} = \frac{483.7}{3600} \approx 0.134 \text{ kWh}Etotal​=3600483.7​≈0.134 kWh

So you need about 0.13–0.14 kWh of energy delivered as heat to raise:

• 2.2 L of water and

• the 15 m of 15 mm copper pipe

from 10 °C to 55 °C.

5.2 Monetary cost at $0.28/kWh

Costtotal=0.134×0.28≈0.0376 dollars\text{Cost}_{\text{total}} = 0.134 \times 0.28 \approx 0.0376 \text{ dollars}Costtotal​=0.134×0.28≈0.0376 dollars

That is approximately:

• $0.038 per heating

• ≈ 3.8 cents (New Zealand dollars if we assume NZ context)

If you ignored the copper and only heated the water volume, the cost would be about 3.2 cents; including the pipe’s thermal mass lifts it to around 3.8 cents.

6. Notes on losses and realism

1. Ambient air at 10 °C, no insulation

   • During the heating process, some heat will leak out through the copper to the air.

   • Our calculation effectively assumes a quick heat-up where those transient losses are small compared with the total energy added. For a slow, low-power heater or very long heat-up times, actual energy use would be a bit higher.

2. Heater efficiency

   • If it is, for example, a 90 % efficient gas or electric system, the input energy required is:

     Einput=Etotalη=0.1340.9≈0.149 kWhE_{\text{input}} = \frac{E_{\text{total}}}{\eta} = \frac{0.134}{0.9} \approx 0.149 \text{ kWh}Einput​=ηEtotal​​=0.90.134​≈0.149 kWh

     Cost then:

     $$0.149\times 0.28\approx 0.0417\text{ dollars}\approx 4.2\text{ cents}$$

   • For a resistive electric cylinder (nearly 100 % efficient at point of use), the earlier ~3.8 c is a good estimate.

3. If this is a dead-leg / draw-off question

   • Every time that length cools back down to ambient and you re-heat it, you “spend” roughly this amount of energy again.

   • So if you flushed that 15 m section, say, 10 times per day, you would waste on the order of:

     $$10\times 0.038\approx \$0.38\text{ per day}$$

     ≈ $140 per year, ignoring heater inefficiencies and any mixing effects.

Final summary

Supreme Commander, under the stated conditions (15 m of 15 mm copper pipe, stationary water, 10 °C → 55 °C, 28 ¢/kWh, no insulation):

• Energy to heat water + pipe: ≈ 0.13–0.14 kWh

• Cost per full heat-up: ≈ 3.8 c (around 4 c allowing for rounding and heater inefficiency)

• If there are 10 uses of the tap per day each allowing the pipe to cool to 10 degrees. That is $0.40c per day. Multiply by 365 days per year = $146 per year. x 10 years =$1,460 per 10 years. 
• So you can see keeping tanks near sinks, reducing the pipe size and insulating well is worth it.