Answer:
the melting point T = 125.36°C
Step-by-step explanation:
Given that:
The resistance of a platinum thermometer at 0°C is
= 160.0 ohms
The resistance of a platinum thermometer when immersed in a crucible containing a melting substance
= 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:
![R_(20) = R_o(1 + \alpha \Delta T)](https://img.qammunity.org/2021/formulas/chemistry/high-school/1qmf1fsucgaeuni4h1vwcezs1i7qtqsph0.png)
![R_(20) = 160(1 + (0.00392 * (20-0)^0C))](https://img.qammunity.org/2021/formulas/chemistry/high-school/k53uvgxauogrvlqm0t3ixhp34yavphma5z.png)
![R_(20) = 160(1 + (0.00392 * (20))](https://img.qammunity.org/2021/formulas/chemistry/high-school/msw31v1ly4q55o9uy664es5qzqxd04kq4n.png)
![R_(20) = 160(1 + (0.0784)](https://img.qammunity.org/2021/formulas/chemistry/high-school/msxekmhv5zz4h4b00ztbbmqs3oqvmc6fzq.png)
![R_(20) = 160(1.0784)](https://img.qammunity.org/2021/formulas/chemistry/high-school/5hb12fpz8s0wse2imqo18rd9zof2xs5174.png)
![R_(20) = 172.544 \ ohms](https://img.qammunity.org/2021/formulas/chemistry/high-school/xpgawg0h7wcti4nd6zvgucihmbr1erlj6h.png)
The resistance of the platinum thermometer at t°C ,
=
![R_(20)(1 + \alpha \Delta T)](https://img.qammunity.org/2021/formulas/chemistry/high-school/3lpea89ljz1426x1s5f8z1siseffr7hsdh.png)
![243.8 = 172.544(1 + 0.00392 * (T-20)^0C}](https://img.qammunity.org/2021/formulas/chemistry/high-school/wk455l56bx4df9r2qinqdjmnfkaf5ovano.png)
![(243.8)/( 172.544 )= (1 + 0.00392 * (T-20)^0C}](https://img.qammunity.org/2021/formulas/chemistry/high-school/5ca7f2vlmddvwy7iqpv324zwmf7hfvaxet.png)
![1.413 = (1 + 0.00392 * (T-20)^0C}](https://img.qammunity.org/2021/formulas/chemistry/high-school/ash6vga3egpt6l6s0jf15r8sgkonqf8s3g.png)
![1.413-1 = 0.00392 * (T-20)^0C}](https://img.qammunity.org/2021/formulas/chemistry/high-school/hyk0s7yod2m291x5clvufuiexkk6eoxwwx.png)
![0.413 = 0.00392 * (T-20)^0C}](https://img.qammunity.org/2021/formulas/chemistry/high-school/b7q9bbuk82rtjkk5e0npvl83uz0afiblkr.png)
![(0.413 )/(0.00392) = (T-20)^0C}](https://img.qammunity.org/2021/formulas/chemistry/high-school/pfmbqt1p1r4c3uxh70ry840qgli9lhuw9s.png)
105.36°C = (T - 20) °C
T = 105.36°C + 20 °C
T = 125.36°C