Question

use the formula for specific heat to complete the following problems. q = mcδt c = q/(mδt) q = heat added (j) m = mass (g) c = specific heat capacity (j/g°c) δt = change in temperature (°c) 1 a total of 108 joules of heat are absorbed as 116.6 g of lead is heated from 15 0°c to 45 0°c. from these data, what is the specific heat of lead?

Explanation:

Step1: Identify given values

$Q = 108\ J$, $m=116.6\ g$, $T_1 = 15.0^{\circ}C$, $T_2=45.0^{\circ}C$

Step2: Calculate $\Delta T$

$\Delta T=T_2 - T_1=45.0 - 15.0=30.0^{\circ}C$

Step3: Use specific - heat formula

$c=\frac{Q}{m\Delta T}=\frac{108}{116.6\times30.0}$

Step4: Calculate the value of $c$

$c=\frac{108}{3498}\approx0.031\ J/(g^{\circ}C)$

Answer:

$0.031\ J/(g^{\circ}C)$