3. a 12oz can of soft drink (assume m = 340 g) at 25°c is placed in a freezer where the temperature is - 12°c. how much energy must be removed from the soft drink for it to reach this temperature?

Question

Accepted Answer

# Explanation:
## Step1: Determine the specific heat capacity of the soft drink
Assume the soft drink has the same specific heat capacity as water, $c = 4.18\,\text{J/g}\cdot^{\circ}\text{C}$

## Step2: Calculate the temperature change $\Delta T$
$\Delta T=T_{final}-T_{initial}=- 12^{\circ}\text{C}-25^{\circ}\text{C}=-37^{\circ}\text{C}$

## Step3: Use the heat - transfer formula $Q = mc\Delta T$
Substitute $m = 340\,\text{g}$, $c = 4.18\,\text{J/g}\cdot^{\circ}\text{C}$, and $\Delta T=-37^{\circ}\text{C}$ into the formula:
$$
\begin{align*}
Q&=(340\,\text{g})\times(4.18\,\text{J/g}\cdot^{\circ}\text{C})\times(- 37^{\circ}\text{C})\
&=340\times4.18\times(-37)\,\text{J}\
&=(340\times4.18)\times(-37)\,\text{J}\
&=1421.2\times(-37)\,\text{J}\
&=- 52584.4\,\text{J}
\end{align*}
$$

The negative sign indicates that heat is removed from the soft drink.

# Answer:
$52584.4\,\text{J}$

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Question

Explanation:

Step1: Determine the specific heat capacity of the soft drink

Step2: Calculate the temperature change \(\Delta T\)

Step3: Use the heat - transfer formula \(Q = mc\Delta T\)

Answer: