Question

3.39 use the heat equation to calculate the energy for each of the following (see table 3.11 ): a. calories to heat 8.5 g of water from 15 °c to 36 °c b. joules lost when 25 g of water cools from 86 °c to 61 °c c. kilocalories to heat 150 g of water from 15 °c to 77 °c d. kilojoules to heat 175 g of copper from 28 °c to 188 °c

Explanation:

Step1: Recall the heat - equation

The heat - equation is $q = mc\Delta T$, where $q$ is the heat energy, $m$ is the mass of the substance, $c$ is the specific heat capacity, and $\Delta T=T_{final}-T_{initial}$. The specific heat capacity of water $c_{water}=1\ cal/g^{\circ}C$ or $4.184\ J/g^{\circ}C$, and the specific heat capacity of copper $c_{copper}=0.385\ J/g^{\circ}C$.

Step2: Calculate part a

For water, $m = 8.5\ g$, $c = 1\ cal/g^{\circ}C$, $\Delta T=36^{\circ}C - 15^{\circ}C=21^{\circ}C$.
$q=mc\Delta T=(8.5\ g)\times(1\ cal/g^{\circ}C)\times(21^{\circ}C)=178.5\ cal$

Step3: Calculate part b

For water, $m = 25\ g$, $c = 4.184\ J/g^{\circ}C$, $\Delta T=61^{\circ}C - 86^{\circ}C=- 25^{\circ}C$.
$q=mc\Delta T=(25\ g)\times(4.184\ J/g^{\circ}C)\times(-25^{\circ}C)=-2615\ J$ (the negative sign indicates heat loss)

Step4: Calculate part c

For water, $m = 150\ g$, $c = 1\ cal/g^{\circ}C$, $\Delta T=77^{\circ}C - 15^{\circ}C = 62^{\circ}C$.
$q=mc\Delta T=(150\ g)\times(1\ cal/g^{\circ}C)\times(62^{\circ}C)=9300\ cal$. Convert to kilocalories: $q = 9300\ cal\div1000 = 9.3\ kcal$

Step5: Calculate part d

For copper, $m = 175\ g$, $c = 0.385\ J/g^{\circ}C$, $\Delta T=188^{\circ}C - 28^{\circ}C=160^{\circ}C$.
$q=mc\Delta T=(175\ g)\times(0.385\ J/g^{\circ}C)\times(160^{\circ}C)=10780\ J$. Convert to kilojoules: $q = 10780\ J\div1000 = 10.78\ kJ$

Answer:

a. $178.5\ cal$
b. $- 2615\ J$
c. $9.3\ kcal$
d. $10.78\ kJ$