# Just a simple question regarding measurements

• Nov 21st 2009, 06:10 AM
meddi83
Just a simple question regarding measurements
Hi guys,

I am given a function C=a + b*T + c*T^2 +d*T^3 which is measured in [KJ/(Kg*K)]

I am given another function h(T) = Integral(0 -> T): CdT and the measuremeant of h(T) is [KJ/Kg].

A value Ta is given, in Celcius, so I converted it to Kelvin (Ta' = Ta+273).

I then found the integral of h(Ta') which is: a*Ta' + (b*Ta'^2)/2 + (c*Ta'^3)/3 + (d*Ta'^4)/4

Is the answer I found in [KJ/Kg] or [KJ/(Kg*K)] ?

If it's the latter, how do I make it in [KJ/Kg] ?

Regards
• Nov 21st 2009, 06:18 AM
skeeter
Quote:

Originally Posted by meddi83
Hi guys,

I am given a function C=a + b*T + c*T^2 +d*T^3 which is measured in [KJ/(Kg*K)]

I am given another function h(T) = Integral(0 -> T): CdT and the measuremeant of h(T) is [KJ/Kg].

A value Ta is given, in Celcius, so I converted it to Kelvin (Ta' = Ta+273).

I then found the integral of h(Ta') which is: a*Ta' + (b*Ta'^2)/2 + (c*Ta'^3)/3 + (d*Ta'^4)/4

Is the answer I found in [KJ/Kg] or [KJ/(Kg*K)] ?

If it's the latter, how do I make it in [KJ/Kg] ?

Regards

$dT$ is in degrees $K$ ...

$\int_0^T C(T) \, dT = \frac{kJ}{kg \cdot K} \cdot K = \frac{kJ}{kg}$
• Nov 21st 2009, 06:24 AM
meddi83
Thank you mate :-)

By the way is this the right sub-forum to ask questions regarding Bisection/Newton-Raphson/Single-point root-estimating methods ?