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 ______ \(^{\circ}\text{C}\).
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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.
Step 1: Set up the ratio form of the radiation law.
A radiation pyrometer reads a fixed radiant energy $E$ off the surface and turns it into a temperature through $E=\varepsilon\sigma T^4$. Since the $\sigma$ and the measured $E$ are the same in both the pyrometer's calculation and reality, we can write
\[
\left(\frac{T_{true}}{T_{calc}}\right)^4 = \frac{\varepsilon_{assumed}}{\varepsilon_{true}}
\]
This comes directly from dividing $\varepsilon_{true}\sigma T_{true}^4=\varepsilon_{assumed}\sigma T_{calc}^4$ on both sides by $\varepsilon_{true}\sigma T_{calc}^4$. Step 2: Convert the given reading to kelvin.
$T_{calc}=1000^{\circ}\text{C}=1273\ \text{K}$. All power-law temperature relations only hold on an absolute scale, so this conversion must happen before any ratio is taken. Step 3: Take the fourth root of the emissivity ratio.
\[
\frac{T_{true}}{T_{calc}} = \left(\frac{0.8}{0.7}\right)^{1/4} = (1.1429)^{0.25} = 1.0339
\] Step 4: Scale up the kelvin reading.
\[
T_{true} = 1273\times1.0339 \approx 1316.2\ \text{K}
\] Step 5: Convert back to Celsius and match with the options.
\[
T_{true} = 1316.2-273 = 1043.2^{\circ}\text{C}
\]
This rounds to $1043^{\circ}\text{C}$, which is option (A). Checking the other three: taking the ratio upside down gives about $958^{\circ}\text{C}$ (option B); skipping the kelvin conversion and scaling $1000$ directly by $1.0339$ gives about $1033^{\circ}\text{C}$ (option C); doing both mistakes together lands near $967^{\circ}\text{C}$ (option D). None of these respects both the absolute temperature scale and the correct ratio direction, so they fall away as arithmetic slips rather than valid readings of the physics.
\[
\boxed{T_{true}\approx1043^{\circ}\text{C}}
\]
