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A thermometer has a mercury thread of length 18 mm at 0°C and 138 mm at 100°C. What is the temperature of the liquid if the mercury thread is 118 mm long?

Physics

ScanSolve AI Answer

Answer

83.33∘C83.33^\circ C

Step-by-step solution

3 steps

To calculate the temperature of the liquid using an uncalibrated thermometer, we use the linear relationship between the length of the mercury thread and the temperature.

  1. Identify the given values.

    Length at melting ice (0∘C0^\circ C), L0=18 mmL_0 = 18\, \text{mm}
    Length at steam (100∘C100^\circ C), L100=138 mmL_{100} = 138\, \text{mm}
    Length at unknown temperature (TT), LT=118 mmL_T = 118\, \text{mm}

  2. Apply the formula for temperature conversion.

    The temperature TT can be calculated using the formula:

    T=LT−L0L100−L0×100∘CT = \frac{L_T - L_0}{L_{100} - L_0} \times 100^\circ C
  3. Substitute the values into the formula and calculate.

    T=118 mm−18 mm138 mm−18 mm×100∘CT = \frac{118\, \text{mm} - 18\, \text{mm}}{138\, \text{mm} - 18\, \text{mm}} \times 100^\circ C T=100 mm120 mm×100∘CT = \frac{100\, \text{mm}}{120\, \text{mm}} \times 100^\circ C T=1012×100∘CT = \frac{10}{12} \times 100^\circ C T=56×100∘CT = \frac{5}{6} \times 100^\circ C T=5006∘CT = \frac{500}{6}^\circ C T=2503∘CT = \frac{250}{3}^\circ C T≈83.33∘CT \approx 83.33^\circ C

    The temperature of the liquid is 83.33∘C83.33^\circ C.

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