---
title: "Plate with convection and radiation: inside temperature"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 7
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:plate-convection-radiation-inside-temp"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/plate-convection-radiation-inside-temp
site: BCT Engineering
---

# Plate with convection and radiation: inside temperature

Fundamental of Thermodynamics & Heat Transfer · Chapter 7: Introduction to Heat Transfer

- Asked in: 2071 Bhadra, 6 marks
- Video: [ME 220 Example problem - convection & radiation of heat from a computer chip](https://www.youtube.com/watch?v=HO4sAh4tuho) — Prof. Birdwell
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/plate-convection-radiation-inside-temp
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-7

## Question

A 2.5 cm thick plate (k = 50 W/mK) 50 cm by 75 cm is maintained at 300°C. Heat is lost from the plate surface by convection and radiation to the ambient air at 20°C. If the emissivity of the surface is 0.9 and the convection heat transfer coefficient is 20 W/m²K, determine the inside plate temperature. [σ = 5.67 × 10⁻⁸ W/m²K⁴]

## Short answer

Heat flux from the surface: convection q_conv = 20×(300−20) = 5600 W/m², radiation q_rad = 0.9×5.67e-8×(573.15⁴−293.15⁴) = 5130 W/m², total q = 10730 W/m². Conduction temperature drop ΔT = qL/k = 10730×0.025/50 = 5.36 K. Inside plate temperature = 300 + 5.36 ≈ 305.4 °C.

## Answer

**Reading of the data.** The 300 °C is taken as the temperature of the exposed outer surface, from which heat leaves by convection and radiation. The inside (other) face of the plate is hotter, because heat reaches the surface by conduction through the plate. Heat flow is steady and one-dimensional.

**Given:** L = 2.5 cm = 0.025 m, k = 50 W/m·K, A = 0.5 × 0.75 = 0.375 m², T_s = 300 °C = 573.15 K, T_∞ = 20 °C = 293.15 K, ε = 0.9, h = 20 W/m²·K, σ = 5.67×10⁻⁸ W/m²·K⁴.

#### Heat flux leaving the surface
Convection: q_conv = h (T_s − T_∞) = 20 × 280 = 5600 W/m²

Radiation (to surroundings at the air temperature):
q_rad = εσ (T_s⁴ − T_∞⁴) = 0.9 × 5.67×10⁻⁸ × (573.15⁴ − 293.15⁴)
= 0.9 × 5.67×10⁻⁸ × (1.0791×10¹¹ − 0.0739×10¹¹)
= 5130 W/m²

Total: q = 5600 + 5130 = 10730 W/m²

Heat flow rate: Q = qA = 10730 × 0.375 = 4024 W

#### Temperature drop across the plate
Conduction: q = k (T_i − T_s)/L

T_i − T_s = qL/k = 10730 × 0.025/50 = 5.36 K

#### Inside temperature
T_i = 300 + 5.36

**Answer: inside plate temperature T_i ≈ 305.4 °C.**

## Check yourself

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

1. The 300 °C is taken as
   A. The exposed outer surface temperature
   B. The ambient air temperature
   C. The mean plate temperature
   D. The inside face temperature
   Answer: A. Heat leaves this surface by convection and radiation. (Key point: Surface temperature)

2. In the radiation term, temperatures must be in
   A. Fahrenheit
   B. Kelvin
   C. Either
   D. Celsius
   Answer: B. q_rad = εσ(Ts^4 - T∞^4) uses absolute temperatures. (Key point: Radiation absolute)

3. Heat flux leaving the surface is
   A. Radiation only
   B. Convection only = 5600
   C. Convection minus radiation
   D. Convection plus radiation = 10730 W/m^2
   Answer: D. 5600 + 5130 = 10730 W/m^2. (Key point: Total flux)

4. Inside temperature is
   A. qL/k = 5.4
   B. Ts = 300
   C. Ts - qL/k
   D. Ts + qL/k = 305.4 °C
   Answer: D. Heat flows inward-to-outward by conduction, so inside is hotter by 5.36 K. (Key point: Inside temperature)

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