a sample of an unknown substance has a mass of 0.465 kg. if 3,000.0 j of heat is required to heat the…

a sample of an unknown substance has a mass of 0.465 kg. if 3,000.0 j of heat is required to heat the substance from 50.0°c to 100.0°c, what is the specific heat of the substance? use $q = mc_{p}delta t$.\n0.00775 j/(g°c)\n0.0600 j/(g°c)\n0.129 j/(g°c)\n0.155 j/(g°c)

a sample of an unknown substance has a mass of 0.465 kg. if 3,000.0 j of heat is required to heat the substance from 50.0°c to 100.0°c, what is the specific heat of the substance? use $q = mc_{p}delta t$.\n0.00775 j/(g°c)\n0.0600 j/(g°c)\n0.129 j/(g°c)\n0.155 j/(g°c)

Answer

Answer:

C. $0.129\ J/(g\cdot ^{\circ}C)$

Explanation:

Step1: Convertir masa a gramos

$m = 0.465\ kg\times1000\ g/kg = 465\ g$

Step2: Calcular el cambio de temperatura

$\Delta T=T_f - T_i=100.0^{\circ}C - 50.0^{\circ}C = 50.0^{\circ}C$

Step3: Reorganizar la fórmula para $C_p$

$C_p=\frac{q}{m\Delta T}$

Step4: Sustituir valores

$C_p=\frac{3000.0\ J}{465\ g\times50.0^{\circ}C}\approx0.129\ J/(g\cdot ^{\circ}C)$