1. Note the correct spelling of celsius.
2. It won't make any difference at all. You're working with this equation.
q = mass x specific heat x (Tfinal-Tinitial).
As an example, let's say a REAL thermometer reads Tfinal as 50 C and Tinitial as 25 C, then delta T is 50-25 = 25. Now let's use the mis-calibrated thermometer(assuming it's mis-calibrated by the same amount for the range being measured) and it's two degrees too high. So it will read 52 for Tf and 27 for Ti. 52-27 = 25 so delta T stays the same and q will not be affected.
What is the effect of plus two degree celcius miscalibrated themometer on the enthalpy of neutralization?
1 answer
1 answer