what is the specific heat of a substance if a mass of 10.0 kg increases in temperature from 10.0°c to 70.0°c…

what is the specific heat of a substance if a mass of 10.0 kg increases in temperature from 10.0°c to 70.0°c when 2,520 j of heat is applied? use $q = mc_{p}delta t$.\n0.00420 j/(g°c)\n0.00661 j/(g°c)\n238 j/(g°c)\n252 j/(g°c)
Answer
Answer:
A. $0.00420\ J/(g\cdot ^{\circ}C)$
Explanation:
Step1: Convert masa a gramos
$m = 10.0\ kg\times1000 = 10000\ g$
Step2: Calcular el cambio de temperatura
$\Delta T=T_{final}-T_{inicial}=70.0^{\circ}C - 10.0^{\circ}C=60.0^{\circ}C$
Step3: Re - organizar la fórmula para $C_p$
$C_p=\frac{q}{m\Delta T}$
Step4: Sustituir valores
$C_p=\frac{2520\ J}{10000\ g\times60.0^{\circ}C}= 0.00420\ J/(g\cdot ^{\circ}C)$