Question

A thermometer gives a reading of \(96.1^{\circ} \mathrm{F} \pm 0.2^{\circ} \mathrm{F}\). What is the temperature in \({ }^{\circ} \mathrm{C}\) ? What is the uncertainty?

Short answer

The temperature in Celsius is approximately \(35.61^{\circ}\mathrm{C}\) with an uncertainty of ±\(0.11^{\circ}\mathrm{C}\).

Step-by-step solution

Identify the conversion formula

We will use the following formula to convert the temperature from Fahrenheit (°F) to Celsius (°C): \( T_{C} = \frac{5}{9}(T_{F} - 32) \) Where: - \(T_{C}\) is the temperature in Celsius - \(T_{F}\) is the temperature in Fahrenheit

Convert the given temperature

Given a temperature reading of \(96.1^{\circ}\mathrm{F}\), we can substitute this value into the conversion formula mentioned above: \( T_{C} = \frac{5}{9}(96.1 - 32)\) Now, solve for the temperature in Celsius: \( T_{C} = \frac{5}{9}(64.1)\) \( T_{C} \approx 35.61^{\circ}\mathrm{C} \) Thus, the temperature is approximately \(35.61^{\circ}\mathrm{C}\) in Celsius.

Calculate the uncertainty

The uncertainty in the Fahrenheit reading is \(0.2^{\circ} \mathrm{F}\). To find the uncertainty in the Celsius reading, we differentiate the conversion formula with respect to temperature in Fahrenheit and multiply it by the uncertainty in Fahrenheit. Differentiating the conversion formula with respect to temperature in Fahrenheit: \( \frac{dT_{C}}{dT_{F}} = \frac{5}{9} \) Now, multiply this by the uncertainty in Fahrenheit: \( \Delta T_{C} = \frac{5}{9} \times 0.2^{\circ} \mathrm{F} \) \( \Delta T_{C} \approx 0.11^{\circ} \mathrm{C} \) So the uncertainty in the Celsius reading is approximately \(0.11^{\circ} \mathrm{C}\).

Write the final answer

The temperature in Celsius is: \( 35.61^{\circ} \mathrm{C}\pm 0.11^{\circ} \mathrm{C} \) Hence, the converted temperature is approximately \(35.61^{\circ}\mathrm{C}\) with an uncertainty of ±\(0.11^{\circ}\mathrm{C}\).