Question

The specific heat of a substance varies with temperature according to c = 0.68 + 0.82T + 0.04072, with T in degrees C and c in cal/g middot K. Find the energy (in cal) required to raise the temperature of 6.5 g of this substance from 4.0 degrees C to 26 degrees C. Number Units

Answer 1

Q = integration m*c*dT

Q = m*[(0.68*(t2-t1) ) + (0.82*(t2^2-t1^2)/2) + (0.04*(t2^3-t1^3)/3)]

Q = 2839.2 calories

Answer 2

let dQ heat increases the temperature by dT

then, $dQ=6.5*(0.68+0.82T+0.040T^{2})dT$

integrating this for total value of heat given:

$Q=\int dQ=\int_{4}^{26}6.5*(0.68+0.82T+0.040T^{2})dT$

or,

$Q=[6.5*(0.68T+0.82T^{2}/2+0.040T^{3}/3)]_{4}^{26}$

or Q=3374 cal=3.374 K calorie

Answer 3

$Q=\int_{T_{1} }^{T _{2}}c.m.dT =6.5\int_{4}^{26}(0.68+0.82T+0.04T^{2})dT$

$Q=6.5[0.68T+0.41T^{2}+0.0133T^{3}]^{26}_{4}=6.5[17.68+277.16+234.29-2.72-6.56-0.8533]$