Question

The readings, tA and tB, of two Celsius thermometers, A and B, agree at ice-point (0 oC) and steam point (100 oC), but elsewhere are related by equation tA = l + m tB + n tB 2 where l, m, and n are constants. When both thermometers are immersed in well-stirred oil bath, A registers 51 oC while B registers 50 oC. Determine the reading on B when A reads 25 oC

Answer 1

Answer:

Explanation:

From the case of well-stirred oil bath:

$51=l+50m+n(50^{2})\rightarrow 51=l+50m+2500n$

At the ice point, both of the thermometers show the same scale:

$0 = l + m(0) + n(0^{2}) \rightarrow l = 0$

At the steam point, again, both of the thermometers show the same scale:

$100 = 0 + m(100)+n(100^{2}) \rightarrow 100 = 100m + 10000n \rightarrow 1 = m + 100n$

By eliminating those equations, we find:

$51=50(1-100n)+2500n \rightarrow 51=50-500n+2500n \rightarrow 1 = 2000n$

so we can obtain that: $n=\frac{1}{2000}=0.0005$ and $m=1-100(0.0005)=1-0.05=0.95$

Now, we have the complete description of the relation between A and B scale as: $t_{A}=0.95t_{B}+0.0005t_{B}^{2}$

So, for $t_{A}=25^{0}C$:

$25 = 0.95t_{B}+0.0005t_{B}^{2} \rightarrow 0.0005t_{B}^{2}+0.95t_{B}-25=0$

$t_{B}_{1,2}=\frac{-0.95\pm\sqrt{0.95^{2}-4(0.0005)(-25)}}{2(0.0005)}\approx-950\pm975$

$t_{B}_{1}=25^{0}C \vee t_{B}_{2}=-1925^{0}C$