"For a first order instrument with a sensitivity of .4 mV/K and a time" constant of 25 ms, find the instrument’s response as a f

Assuming: For a first-order instrument with a sensitivity of 0.4 mV/K and a time constant of 25 ms, the initial temperature is 273 K and the final temperature is 473 K. Initially, the volume is 0.4 mV/K multiplied by 273 K, equating to 109.2 V. The final volume is calculated as 0.4 mV/K multiplied by 473 K, resulting in 189.2 V. The response of the instrument over time for a rapid rise in temperature is as follows: Considering y as the function of time, we have y(t) = 109.2 + (189.2 - 109.2)(1 - ) mV. This simplifies to y(t) = (109.2 + 80(1 - )) mV. You can graph the response y(t) over time. The response graph of y(t) over time is displayed in the diagram below. The unit for y(t) is mV. To determine the 90% rise time for y(t90) along with the error fraction, you set 90% of 189.2 mV, thus: 0.9 × 189.2 mV = 170.28 mV. When rewritten, we get 170.28 mV = (109.2 + 80(1 - )) mV. This leads to 170.28 mV - 109.2 mV = 80(1 - )) mV, yielding 61.08 mV as 80(1 - )) mV. Thus, 0.7635 mV = (1 - )) mV. When calculating time, we find that t = 1.44 × 25 × 10⁻³ s = 0.036 s, which converts to 36 ms. Thus, the error fraction works out to be 0.1 or 10%.