A student measures a boiling water bath with two different thermometers. The digital thermometer records the temperature as 100.2°C and has an uncertainty of 0.1°C. The analog thermometer records the temperature as 99.0°C and has an uncertainty of 0.5°C. What is the percent uncertainty of each thermometer? And what is the percent error given that water boils at 100.0°C?

Explanation:

Measurement made = 100.2 °C

uncertainty = 0.1°C

percent uncertainty = .1 x 100 / 100.2

= .099 %

2 nd thermometer :

Measurement made = 99.0 °C

uncertainty = 0.5°C

percent uncertainty = .5 x 100 / 99

= .505  %

Percent error :

1 st thermometer

Error = 100.2 - 100 = .2 °C

measurement = 100.2

percent error = .2 x 100 / 100.2

= .1996 %

2 nd thermometer

Error = 100.00 - 99 = 1.00 °C

measurement = 99

percent error = 1 x 100 / 99

= 1.01  %

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.