Thermodynamic quantity equivalent to the total heat content of a system It is equal to the internal energy of the system plus the product of pressure and volume
The absorbance reported by the defective instrument was 0.3933.
Absorbance A = - log₁₀ T
Tm = transmittance measured by spectrophotometer
Tm = 0.44
Absorbance reported in this equipment = -log₁₀ (0.44) = 0.35654
True absorbance can be calculated by true transmittance, Tm = T+S(α-T)
S = fraction of stray light = 6%= 6/100 = 0.06
α= 1, ideal case
T = true transmittance of the sample
Tm = T+S(α-T)
now, T= Tm-S/ 1-S = 0.44-0.06/ 1-0.06 = 0.404233
therefore, actual reading measured is A = -log₁₀ T = -log₁₀ (0.404233)
i.e; 0.3933
To know more about transmittance click here:
brainly.com/question/17088180
#SPJ4
Answer:
The corect answer would be C.
Explanation:
The flow rate set at a differentt time would be the correct measurement beecause wats and speed add up to your main answer.
<span>If I done the math correctly it is 3729J because you multiply 16.5 g by the 2260 J/g and get 3729 J</span>
Answer and Explanation:
Calorie is the unit of heat energy . There are 2 units with the same name 'calorie' which is widely used.
'The amount of heat energy required to increase the temperature of 1 gram of water by mass by 
or 1 K is known as small calorie or gram calorie'.
Another one is large calorie which can be defined as :
'The amount of heat energy required to make arise in temperature of water 1 kg by mass by or 1 K is known as large calorie or kilcalorie and is represented as Cal or kcal'.
After the adoption of SI system, thee units of the metric system cal, C or kilocal are considered deprecated or obsolete with the SI unit for heat energy as 'joule or J'
1 cal = 4.184 J
1C or 1 kilocal = 4184 J
Calorimeter constant:
Calorimeter constant, represented as '
' is used to quantify the heat capacity or the amount of heat of a calorimeter.
It can be calculated by ther given formula:
where,
= corresponding temperature change
= enthalpy change
Its unit is J/K or J/1^{\circ}C[/tex] which can be convertyed to cal/1^{\circ}C[/tex] by dividing the calorimeter constant by 4.184 or 4184 accordingly.
