Question

6,340 joules of energy were absorbed by the 245.0 g piece of lithium when it was placed in the calorimeter and allowed to reach equilibrium at 52.0 °c. what was the initial temperature of the lithium?
$c_{lithium} = 3.57$ j/g °c
$t_i = ?$ °c

Explanation:

Step1: Recall the heat formula

The formula for heat absorbed or released is \( Q = mc\Delta T \), where \( Q \) is heat, \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T = T_f - T_i \) (change in temperature, final minus initial). Here, \( Q = 6340 \, \text{J} \), \( m = 245.0 \, \text{g} \), \( c = 3.57 \, \text{J/g}^\circ\text{C} \), \( T_f = 52.0^\circ\text{C} \), and we need to find \( T_i \).

Step2: Rearrange the formula to solve for \( \Delta T \)

From \( Q = mc\Delta T \), we can solve for \( \Delta T \): \( \Delta T = \frac{Q}{mc} \)

Step3: Calculate \( \Delta T \)

Substitute the values: \( \Delta T = \frac{6340}{245.0 \times 3.57} \)
First, calculate the denominator: \( 245.0 \times 3.57 = 245.0 \times 3.57 = 874.65 \)
Then, \( \Delta T = \frac{6340}{874.65} \approx 7.25 \) (rounded to two decimal places)

Step4: Relate \( \Delta T \) to \( T_i \) and \( T_f \)

We know that \( \Delta T = T_f - T_i \), so we can rearrange to solve for \( T_i \): \( T_i = T_f - \Delta T \)

Step5: Calculate \( T_i \)

Substitute \( T_f = 52.0^\circ\text{C} \) and \( \Delta T \approx 7.25^\circ\text{C} \): \( T_i = 52.0 - 7.25 = 44.75^\circ\text{C} \) (Wait, no, wait. Wait, if the energy is absorbed, that means the temperature of lithium increased, so \( T_f > T_i \), so \( \Delta T = T_f - T_i \), so \( T_i = T_f - \Delta T \)? Wait, no, wait. Wait, if the lithium absorbed energy, its temperature should increase, so \( T_f = T_i + \Delta T \), so \( T_i = T_f - \Delta T \)? Wait, no, let's check the formula again. The heat absorbed is \( Q = mc(T_f - T_i) \) because the final temperature is higher than initial when absorbing heat. So we have \( 6340 = 245.0 \times 3.57 \times (52.0 - T_i) \)? Wait, no, wait, that can't be, because if \( T_f - T_i \) is positive, then \( T_f > T_i \), which is correct for absorbing heat (temperature increases). Wait, but when we calculated \( \Delta T = Q/(mc) = 6340/(245*3.57) \approx 6340/874.65 \approx 7.25 \). So \( 52.0 - T_i = 7.25 \), so \( T_i = 52.0 - 7.25 = 44.75 \)? Wait, but that would mean the temperature decreased, which contradicts the absorption of energy. Wait, I must have messed up the sign. Wait, no, the formula is \( Q = mc\Delta T \), where \( \Delta T = T_f - T_i \) when heat is absorbed (temperature increases). So if \( Q \) is positive (absorbed), then \( \Delta T \) is positive, so \( T_f > T_i \). So my calculation of \( \Delta T \) was correct as \( 7.25 \), so \( T_i = T_f - \Delta T = 52.0 - 7.25 = 44.75 \)? But that would mean the lithium was at 44.75 and heated to 52, absorbing energy. Wait, but let's recalculate \( \Delta T \):

\( 245.0 \times 3.57 = 245 * 3.57 \). Let's calculate 245 * 3 = 735, 245 * 0.57 = 245 * 0.5 + 245 * 0.07 = 122.5 + 17.15 = 139.65, so total is 735 + 139.65 = 874.65. Then 6340 / 874.65 ≈ 7.25. So \( T_f - T_i = 7.25 \), so \( T_i = T_f - 7.25 = 52.0 - 7.25 = 44.75 \). Wait, but that seems low. Wait, maybe I made a mistake in the formula. Wait, no, the specific heat of lithium is 3.57 J/g°C, which is relatively high. Let's recalculate \( \Delta T \): 6340 divided by (245 * 3.57). 245 * 3.57: 200*3.57=714, 45*3.57=160.65, so total 714+160.65=874.65. Then 6340 / 874.65 ≈ 7.25. So 52 - 7.25 = 44.75. So the initial temperature is approximately 44.8 °C (rounded to three significant figures? Wait, the given values: 6340 J (four significant figures), 245.0 g (four), 3.57 J/g°C (three), 52.0 °C (three). So the least number of significant figures is three, so the answer should be 44.8 °C? Wait, 7.25 is approximately 7.25, 52.0 - 7.25 = 44.75, which rounds to 44.8 with three significant figures. Wait, but let's check the calculation again. Wait, maybe I had the formula reversed. Let's do it step by step.

Given \( Q = mc\Delta T \), \( \Delta T = T_f - T_i \) (since heat is absorbed, final temp is higher). So:

\( 6340 = 245.0 \times 3.57 \times (52.0 - T_i) \)

Wait, no, that's not right. Wait, \( Q = mc(T_f - T_i) \) when the substance is heated (absorbing heat), so \( T_f > T_i \). So solving for \( T_i \):

\( T_f - T_i = \frac{Q}{mc} \)

\( T_i = T_f - \frac{Q}{mc} \)

So plugging in the numbers:

\( T_i = 52.0 - \frac{6340}{245.0 \times 3.57} \)

Calculate \( \frac{6340}{245.0 \times 3.57} \):

245.0 * 3.57 = 874.65

6340 / 874.65 ≈ 7.248

So \( T_i = 52.0 - 7.248 ≈ 44.752 \), which is approximately 44.8 °C (three significant figures) or 44.75 °C (four significant figures). Wait, but let's check if the energy is absorbed, so the lithium's temperature increases, so initial temperature is lower than final. So 44.75 °C is lower than 52.0 °C, so that makes sense. So the initial temperature is approximately 44.8 °C (or 44.75 °C).

Wait, maybe I made a mistake in the sign. Let's confirm with the formula. The heat absorbed by the lithium is equal to \( mc\Delta T \), where \( \Delta T \) is the change in temperature (final - initial). So if the lithium absorbed energy, its temperature went from \( T_i \) to \( T_f = 52.0 \), so \( T_f > T_i \), so \( \Delta T = T_f - T_i \) is positive, and \( Q = mc\Delta T \) is positive, which matches the given \( Q = 6340 \) J (positive, absorbed). So the calculation is correct. So \( T_i = 52.0 - \frac{6340}{245.0 \times 3.57} \approx 52.0 - 7.25 = 44.75 \) °C, which can be rounded to 44.8 °C (three significant figures) or 44.75 °C (four significant figures).

Answer:

The initial temperature of the lithium is approximately \(\boxed{44.8}\) °C (or 44.75 °C, depending on rounding).