---
title: "Resistance spot welding heat and efficiency"
subject: "Workshop Technology"
chapter: 10
key: "semester_2/workshop_technology:resistance-spot-welding-heat-numerical"
source: https://www.bctengineer.com/q/semester_2/workshop_technology/resistance-spot-welding-heat-numerical
site: BCT Engineering
---

# Resistance spot welding heat and efficiency

Workshop Technology · Chapter 10: Metal Joining Process

- Asked in: year not given, 4+4 marks
- Video: [Resistance welding process | Spot welding and Seam welding process](https://www.youtube.com/watch?v=QDayHY-1obg) — ADTW Study
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/workshop_technology/resistance-spot-welding-heat-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_2/workshop_technology/chapter-10

## Question

Explain the principle of resistance spot welding with a sketch and state its applications. In a spot welding operation a current of 9000 A flows for 0.25 s through a joint with a resistance of 120 micro-ohm. The weld nugget has a mass of 0.25 g of steel, with specific heat 0.5 J/g K, melting rise of 1500 K and latent heat of fusion 270 J/g. Find the heat generated and the efficiency of the process.

## Short answer

Resistance spot welding passes high current through overlapping sheets, generating heat at the interface to form a nugget. Applications: car bodies, sheet‑metal assemblies. Heat generated I²Rt = 9000² × 120 × 10⁻ ⁶ × 0.25 = 2430 J. Heat to melt nugget = 0.25 g × (0.5 × 1500 + 270) J/g = 255 J. Efficiency = 10.5 %.

## Answer

#### Principle
Two overlapping sheets are clamped between copper alloy electrodes, and a large current passes for a short time. The joint resistance generates heat by the Joule effect, $H = I^2 R t$, which melts a small nugget at the interface. The pressure is held while the nugget solidifies, forming a spot weld.

```text
   electrode (Cu)   <- force
        |  |
  ======XXXX======  sheets (nugget X)
        |  |
   electrode (Cu)   <- force
```
Steps: squeeze, weld (current on), hold, release.

Applications: car bodies, metal furniture, utensils, electric boxes, wire mesh, thin sheet steel assemblies; high production and easily automated.

#### Heat generated
$$
\begin{aligned} H &= I^2 R\,t \\ &= (9000)^2\times 120\times 10^{-6}\times 0.25 \\ &= 8.1\times 10^{7}\times 1.2\times 10^{-4}\times 0.25 = 2430\ \text{J} \end{aligned}
$$

#### Heat needed for the nugget
Heat to raise 1 g by 1500 K and melt it:

$$
q = c\,\Delta T + L = 0.5\times 1500 + 270 = 1020\ \text{J/g}
$$

$$
H_{nugget} = 0.25\times 1020 = 255\ \text{J}
$$

#### Efficiency
$$
\eta = \frac{255}{2430}\times 100 = 10.5\%
$$

The rest of the heat is lost by conduction into the water-cooled electrodes and the surrounding sheet.

**Answer: heat generated = 2430 J; heat used to melt the nugget = 255 J; efficiency = 10.5%.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is heat produced in resistance spot welding?
   A. By an arc between electrode and sheet
   B. By a gas flame
   C. By Joule heating, H = I^2 R t, at the joint resistance
   D. By friction between sheets
   Answer: C. A large current through the joint resistance melts a small nugget; pressure is held while it solidifies. (Key point: Principle of spot welding)

2. What heat is generated for 9000 A, 120 micro-ohm, 0.25 s?
   A. 2430 J
   B. 270 J
   C. 24,300 J
   D. 255 J
   Answer: A. H = 9000^2 x 120e-6 x 0.25 = 2430 J. (Key point: Heat generated)

3. How is the heat needed for the nugget found?
   A. Only latent heat: 0.25 x 270
   B. Only sensible heat: 0.25 x 0.5 x 1500
   C. Mass x (c ΔT + L) = 0.25 x 1020 = 255 J
   D. Mass x c x L
   Answer: C. The metal must be heated by 1500 K and melted, so both sensible and latent heat count. (Key point: Heat needed for nugget)

4. What is the efficiency and where does the rest of the heat go?
   A. 10.5%; lost by conduction into cooled electrodes and surrounding sheet
   B. 95%; lost as light
   C. 10.5%; stored in the nugget
   D. 50%; lost as sound
   Answer: A. 255/2430 = 10.5%; the remainder conducts into water-cooled electrodes and sheet. (Key point: Efficiency)

## Related in Workshop Technology

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