Question

The emitted radiant energy from a piece of metal was measured using a pyrometer. The temperature was calculated to be \(1000^{\circ}\text{C}\), assuming a surface emissivity of \(0.8\). It was later found that the true surface emissivity was \(0.7\).
The actual temperature of the object is nearest to ______ \(^{\circ}\text{C}\).

• A. \(1043\)
• B. \(958\)
• C. \(1033\)
• D. \(967\)

Hint:

Use \(E=\varepsilon\sigma T^4\) with the same emitted energy on both sides: \(\varepsilon_{true}T_{true}^4=\varepsilon_{assumed}T_{calc}^4\), and remember to work in kelvin.

Answer 1

The Correct Option is A

Step 1: Understand how a radiation pyrometer gets its reading.
A radiation pyrometer looks at a hot surface without touching it and picks up the total radiant energy the surface gives off. It converts that energy into a temperature using the Stefan-Boltzmann law,
\[ E = \varepsilon\,\sigma\,T^4 \]
where \(E\) is the radiant emittance, \(\varepsilon\) is the surface emissivity, \(\sigma\) is the Stefan-Boltzmann constant, and \(T\) is the absolute temperature. The instrument cannot sense emissivity on its own, so the operator must key in an assumed value of \(\varepsilon\) before the reading can be turned into a temperature.
Step 2: Write the energy relation for both cases.
The metal is really giving off one fixed amount of radiant energy \(E\). Written using its true emissivity and true temperature,
\[ E = \varepsilon_{true}\,\sigma\,T_{true}^4,\qquad \varepsilon_{true}=0.7 \]
The same measured \(E\) was fed into the pyrometer, but it used the wrong, assumed emissivity to work out a reading of \(1000^{\circ}\text{C}\),
\[ E = \varepsilon_{assumed}\,\sigma\,T_{calc}^4,\qquad \varepsilon_{assumed}=0.8,\ T_{calc}=1000^{\circ}\text{C} \]
Step 3: Equate the two expressions for E.
Since both sides equal the same physical \(E\), the \(\sigma\) cancels and
\[ \varepsilon_{true}\,T_{true}^4 = \varepsilon_{assumed}\,T_{calc}^4 \]
Solving for the true temperature,
\[ T_{true} = T_{calc}\left(\frac{\varepsilon_{assumed}}{\varepsilon_{true}}\right)^{1/4} \]
Step 4: Convert to kelvin and plug in the numbers.
The power law needs an absolute temperature, so first convert,
\[ T_{calc} = 1000+273 = 1273\ \text{K} \]
Now find the emissivity ratio and its fourth root,
\[ \frac{\varepsilon_{assumed}}{\varepsilon_{true}} = \frac{0.8}{0.7} = 1.1429,\qquad (1.1429)^{1/4}=1.0339 \]
Step 5: Compute the true temperature.
\[ T_{true} = 1273\times1.0339 = 1316.2\ \text{K} \]
Converting back to Celsius,
\[ T_{true} = 1316.2-273 = 1043.2^{\circ}\text{C}\approx1043^{\circ}\text{C} \]
Step 6: Check the direction of the correction makes physical sense.
The true emissivity \(0.7\) is lower than the assumed \(0.8\). A surface with lower emissivity radiates less strongly at a given temperature, so to give off the same measured energy it must actually be hotter than the \(1000^{\circ}\text{C}\) the instrument reported. Since \(\varepsilon_{true}<\varepsilon_{assumed}\), we expect \(T_{true}>T_{calc}\), which already rules out option (B) \(958\) and option (D) \(967\), since both sit below \(1000^{\circ}\text{C}\). Working the numbers confirms option (B) comes from using the emissivity ratio upside down, \((\varepsilon_{true}/\varepsilon_{assumed})^{1/4}\), which pulls the temperature down instead of up.
Step 7: See where option (C) and (D) go wrong.
Option (C), \(1033\), comes from applying the correct ratio \(1.0339\) directly to the Celsius number \(1000\) instead of converting to kelvin first, a common slip since the fourth-power law is only valid on the absolute scale. Option (D), \(967\), repeats that same Celsius-only mistake and also flips the ratio upside down, compounding both errors. Only the full, correct calculation in kelvin gives \(1043^{\circ}\text{C}\).
Final Answer:
The actual temperature of the object is
\[ \boxed{1043^{\circ}\text{C}} \]