Single-pane window: film resistances and the overnight loss

SPH4C Grade 12 Physics (College) · Energy Transformations

For a home-energy project, a student tapes one logging thermometer to the kitchen wall and hangs another outside the back door, then reads the averages over an 8.00 h winter night: 20.0 °C in, −5.0 °C out. The back door's single-glazed pane measures 1.20 m by 0.80 m (0.960 m²) and its glass is 4.0 mm thick with conductivity 0.96 W/(m·K). A building-science table supplies what the glass alone cannot: the still-air film inside clings at RSI 0.12, and the winter-wind film outside at RSI 0.030.

Given
  • A = 0.96 Area of the pane (1.20 m × 0.80 m)
  • d = 4 mmGlass thickness
  • k = 0.96 W/(m·K)Thermal conductivity of glass
  • R_in = 0.12 m²·K/WInside still-air film resistance
  • R_out = 0.03 m²·K/WOutside wind film resistance
  • ΔT = 25 Inside-to-outside difference (20.0 − (−5.0))
  • t = 8 hLength of the night
Determine
  1. (a)the R-value of the glass layer itself
  2. (b)the total R-value of the path, air films included
  3. (c)the steady heat loss through the pane
  4. (d)the energy that walks out through this one window by morning
Step 1 of 4(a) · solve for R-value of the layer

R = L/k for the glass alone: four thin millimetres of a decent conductor is RSI 0.0042 — next to nothing. Anyone who stops here and computes Q = kAΔT/d gets 5760 W, a space heater's output times four, and the absurdity is the lesson: bare Fourier through the glass is not how a window loses heat.

Rearranged for R
R=LkR = \frac{L}{k}
Your values, in your units
R=(4 mm)(0.96 W/(mK))R = \frac{\left( 4\ \text{mm} \right)}{\left( 0.96\ \text{W/(m}{\cdot}\text{K)} \right)}
Answer
R=41.667 cm2C/WR = 41.667\ \text{cm}^{2}{\cdot}^{\circ}\text{C/W}

Carried onward at full precision, not this rounded figure.

Open the R-Value of an Insulation Layer (R = L/k) solver →

Step 2 of 4(b) · solve for Total assembly R-value

Heat must cross three layers in series — still air clinging inside, glass, moving air outside — and series resistances simply add: RSI 0.154 in total. Run the shares: the two invisible films carry 36 parts in 37 of the resistance, the glass one part. A single pane is mostly a frame for its own air films.

Rearranged for R_tot
Rtot=R1+R2+R3R_{tot} = R_1 + R_2 + R_3
41.667 cm²·°C/Wcarried from step 1
Your values, in your units
Rtot=(0.12 RSI (m2K/W))+(0.00416667 RSI (m2K/W))+(0.03 RSI (m2K/W))R_{tot} = \left( 0.12\ \text{RSI (m}^{2}{\cdot}\text{K/W)} \right) + \left( 0.00416667\ \text{RSI (m}^{2}{\cdot}\text{K/W)} \right) + \left( 0.03\ \text{RSI (m}^{2}{\cdot}\text{K/W)} \right)
Answer
Rtot=1.5417 togR_{tot} = 1.5417\ \text{tog}

Carried onward at full precision, not this rounded figure.

Open the Total R-Value of an Assembly solver →

Step 3 of 4(c) · solve for Heat loss rate

Q = A·ΔT/R_tot: about 156 W flows through this one pane all night — a couple of old light bulbs' worth, continuously. The ΔT is 20.0 − (−5.0) = 25.0 K; mishandling the negative sign to get 15 K is the arithmetic slip this scenario is built to catch.

Rearranged for Q
Q˙=AΔTRtot\dot{Q} = \frac{A \, \Delta T}{R_{tot}}
1.5417 togcarried from step 2
Your values, in your units
Q˙=(0.96 m2)(25 C)(0.154167 RSI (m2K/W))\dot{Q} = \frac{\left( 0.96\ \text{m}^{2} \right) \, \left( 25\ \text{C}^{\circ} \right)}{\left( 0.154167\ \text{RSI (m}^{2}{\cdot}\text{K/W)} \right)}
Answer
Q˙=155.68 W\dot{Q} = 155.68\ \text{W}

Carried onward at full precision, not this rounded figure.

Open the Heat Loss Through an Assembly (Q = A·ΔT/R) solver →

Step 4 of 4(d) · solve for Work or energy

Energy is power held for a time: 155.7 W × 28 800 s ≈ 4.48 MJ by dawn, which the utility would bill as 1.25 kWh — from one window, in one night. The 8.00 h enters as seconds; hours fed raw would understate the loss 3600-fold.

Rearranged for W
W=PtW = P t
155.68 Wcarried from step 3
Your values, in your units
W=(155.676 W)(8 h)W = \left( 155.676\ \text{W} \right) \, \left( 8\ \text{h} \right)
Converted to base units
W=(155.676 W)(28,800 s)W = \left( 155.676\ \text{W} \right) \, \left( 28{,}800\ \text{s} \right)
Answer
W=4.4835 MJW = 4.4835\ \text{MJ}

Carried onward at full precision, not this rounded figure.

Open the Power (P = W/t) solver →

Therefore the glass itself resists almost nothing (RSI 0.0042), the air films raise the path to RSI 0.154, the pane bleeds 156 W across the 25.0 K night, and by morning 4.48 MJ — 1.25 kWh — has left through one small window.

Why this order

The chain is ordered to spring a trap and then disarm it. Part (a) computes the honest R-value of the glass and invites the bare-conduction answer — 5760 W through a pane, which would boil a kettle every four minutes — before part (b) adds the two air films and the estimate collapses by a factor of 37 to something a hand on cold glass actually feels. The physics point is that a series path is governed by its largest resistances, and here the largest resistances are layers no ruler can measure: the millimetres of still air the glass traps against itself indoors, and the thinner, wind-scrubbed film outside. That is why the outside film shrinks on a windy night (and the window loses faster), and why the R-value of any bare pane is almost independent of its glass. Steps (c) and (d) are then plain bookkeeping: rate from resistance, energy from rate and the night's length, each unit conversion done where the note says.

The engineering consequence is the entire glazing industry. If the films are the only real resistance, the way to a better window is more films — which is exactly what double glazing is: a sealed gap of still air (RSI ≈ 0.17 by itself) buys more than doubling the glass a hundred times over would. Argon fills, low-e coatings that suppress the radiative bypass, triple glazing — each is another still layer or a quieter parallel path, and each shows up directly in this same three-step arithmetic. The 1.25 kWh answer also scales soberingly: a house with fifteen such panes loses ~19 kWh a night through glass alone, which is why window upgrades headline every energy audit, and why the RSI/R-value language this chain practices is the one Canadian building code and contractor both speak.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.