a sample of an unknown substance has a mass of 0.158 kg. if 2,510.0 j of heat is required to heat the…
a sample of an unknown substance has a mass of 0.158 kg. if 2,510.0 j of heat is required to heat the substance from 32.0°c to 61.0°c, what is the specific heat of the substance? use $q = mc_{p}delta t$.\n0.171 j/(g·°c)\n0.548 j/(g·°c)\n15.9 j/(g·°c)\n86.6 j/(g·°c)
Answer
Explanation:
Step1: Convert mass to grams
$m = 0.158\ kg=0.158\times1000\ g = 158\ g$
Step2: Calculate the temperature change
$\Delta T=T_2 - T_1=61.0^{\circ}C - 32.0^{\circ}C=29.0^{\circ}C$
Step3: Rearrange the heat - formula to solve for specific heat
Given $q = mC_p\Delta T$, then $C_p=\frac{q}{m\Delta T}$. Substitute $q = 2510.0\ J$, $m = 158\ g$ and $\Delta T=29.0^{\circ}C$ into the formula: $C_p=\frac{2510.0\ J}{158\ g\times29.0^{\circ}C}$ $C_p=\frac{2510.0}{4582}\ J/(g\cdot^{\circ}C)\approx0.548\ J/(g\cdot^{\circ}C)$
Answer:
0.548 J/(g,°C)