Question

a 400.000 g liquid sample lost 234.0 cal when it cooled from 66.6 to 55.5 °c. the specific heat of the liquid is numerically 1 and the units are 2.
a. 6.51 b. 8.45 x 10³ c. 1.04 x 10⁶ d. 0.0528 e. 5.28 f. 8.45
g. 1.04 h. cal/g°c i. cal g/°c j. g °c/cal k. cal g °c l. g/°c cal

Explanation:

Step1: Identify the heat - loss formula

The formula for heat transfer is $q = mc\Delta T$, where $q$ is the heat transferred, $m$ is the mass of the substance, $c$ is the specific heat, and $\Delta T$ is the change in temperature. We want to solve for $c$, so $c=\frac{q}{m\Delta T}$.

Step2: Calculate the change in temperature

$\Delta T=T_{final}-T_{initial}=55.5 - 66.6=- 11.1^{\circ}C$. The heat lost $q=-234.0$ cal (negative because heat is lost) and $m = 400.000$ g.

Step3: Substitute values into the formula for specific heat

$c=\frac{-234.0\ cal}{400.000\ g\times(-11.1^{\circ}C)}=\frac{-234.0}{-4440}\ cal/g^{\circ}C\approx0.0528\ cal/g^{\circ}C$

Answer:

1. D. 0.0528
2. H. cal/g$^{\circ}$C