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∘C=5/9(∘F−32){ }^{\circ} \mathrm{C}=5 / 9\left({ }^{\circ} F-32\right) A 20-Ampere Fuse Element Will Open at 203 Degrees Fahrenheit

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∘C=5/9(∘F−32){ }^{\circ} \mathrm{C}=5 / 9\left({ }^{\circ} F-32\right) A 20-ampere fuse element will open at 203 degrees Fahrenheit due to heat from excessive current. At what Celsius temperature will the fuse open?

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