200 garms into kilogram
2 answers:
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Answer:
ΔTmin = 3.72 °C
Explanation:
With a 16-bit ADC, you get a resolution of steps. This means that the ADC will divide the maximum 10V input into 65536 steps:
ΔVmin = 10V / 65536 = 152.59μV
Using the thermocouple sensitiviy we can calculate the smallest temperature change that 152.59μV represents on the ADC:
Answer:
16.6 kJ/°C
Explanation:
given,
Amount of heat absorbed = 45 kJ
initial temperature, T₁ = 25.5°C
final temperature, T₂ = 28.2°C
change in temperature = T₂ - T₁
= 28.2 - 25.5 = 2.7° C
Heat capacity of the object is equal to 16.6 kJ/°C