Answer:
the melting point T = 125.36°C
Explanation:
Given that:
The resistance of a platinum thermometer at 0°C is [tex]R_o[/tex] = 160.0 ohms
The resistance of a platinum thermometer when immersed in a crucible containing a melting substance [tex]R_t[/tex] = 243.8 ohms
The temperature coefficient at room temperature 20°C = ∝ = 0.00392
The objective is to determine the melting point of this substance
To do that ; at 20°C, the resistance of the platinum thermometer can be calculated as follows:
[tex]R_{20} = R_o(1 + \alpha \Delta T)[/tex]
[tex]R_{20} = 160(1 + (0.00392 \times (20-0)^0C))[/tex]
[tex]R_{20} = 160(1 + (0.00392 \times (20))[/tex]
[tex]R_{20} = 160(1 + (0.0784)[/tex]
[tex]R_{20} = 160(1.0784)[/tex]
[tex]R_{20} = 172.544 \ ohms[/tex]
The resistance of the platinum thermometer at t°C , [tex]R_t[/tex] = [tex]R_{20}(1 + \alpha \Delta T)[/tex]
[tex]243.8 = 172.544(1 + 0.00392 \times (T-20)^0C}[/tex]
[tex]\dfrac{243.8}{ 172.544 }= (1 + 0.00392 \times (T-20)^0C}[/tex]
[tex]1.413 = (1 + 0.00392 \times (T-20)^0C}[/tex]
[tex]1.413-1 = 0.00392 \times (T-20)^0C}[/tex]
[tex]0.413 = 0.00392 \times (T-20)^0C}[/tex]
[tex]\dfrac{0.413 }{0.00392} = (T-20)^0C}[/tex]
105.36°C = (T - 20) °C
T = 105.36°C + 20 °C
T = 125.36°C
