Question

1. if the temperature of 34.4 g of ethanol increases from 25 °c to 78.8 °c, how much heat has been absorbed by the ethanol? the specific heat of ethanol is 2.44 j/(g*°c)
2. graphite has a specific heat of 0.709 j/(g*°c). if a 25 gram piece of graphite is cooled from 35 °c to 18 °c, how much energy was lost by the graphite?
3. assuming that coca cola has the same specific heat as water (4.18 j/g°c), calculate the amount of heat transferred when one can (about 350g) is cooled from 25°c to 3°c.

Explanation:

Step1: Recall heat - transfer formula

The formula for heat transfer is $q = mc\Delta T$, where $q$ is the heat transferred, $m$ is the mass, $c$ is the specific heat, and $\Delta T=T_{final}-T_{initial}$.

Step2: Solve for ethanol problem (Question 3)

Given $m = 34.4\ g$, $c = 2.44\ J/(g\cdot^{\circ}C)$, $T_{initial}=25^{\circ}C$, $T_{final}=78.8^{\circ}C$. Then $\Delta T=78.8 - 25=53.8^{\circ}C$.
$q = mc\Delta T=34.4\ g\times2.44\ J/(g\cdot^{\circ}C)\times53.8^{\circ}C$
$q=34.4\times2.44\times53.8\ J\approx4507.7\ J$

Step3: Solve for graphite problem (Question 4)

Given $m = 25\ g$, $c = 0.709\ J/(g\cdot^{\circ}C)$, $T_{initial}=35^{\circ}C$, $T_{final}=18^{\circ}C$. Then $\Delta T=18 - 35=- 17^{\circ}C$.
$q = mc\Delta T=25\ g\times0.709\ J/(g\cdot^{\circ}C)\times(-17^{\circ}C)$
$q=25\times0.709\times(- 17)\ J\approx - 300.0\ J$ (the negative sign indicates heat loss)

Step4: Solve for Coca - Cola problem (Question 5)

Given $m = 350\ g$, $c = 4.18\ J/(g\cdot^{\circ}C)$, $T_{initial}=25^{\circ}C$, $T_{final}=3^{\circ}C$. Then $\Delta T=3 - 25=-22^{\circ}C$.
$q = mc\Delta T=350\ g\times4.18\ J/(g\cdot^{\circ}C)\times(-22^{\circ}C)$
$q=350\times4.18\times(-22)\ J\approx - 32284\ J$ (the negative sign indicates heat loss)

Answer:

Question 3: Approximately $4507.7\ J$
Question 4: Approximately $300.0\ J$ (heat loss)
Question 5: Approximately $32284\ J$ (heat loss)