Question

a chemical reaction takes place inside a flask submerged in a water bath. the water bath contains 5.50 kg of water at 37.7°c. during the reaction 70.9 kj of heat flows out of the flask and into the bath. calculate the new temperature of the water bath. you can assume the specific heat capacity of water under these conditions is 4.18 j·g⁻¹·k⁻¹. be sure your answer has the correct number of significant digits.

Explanation:

Step1: Convert mass to grams and heat to joules

Mass of water \( m = 5.50\space kg = 5.50\times1000 = 5500\space g \)
Heat \( q = 70.9\space kJ = 70.9\times1000 = 70900\space J \)

Step2: Use the heat formula \( q = mc\Delta T \) to find \( \Delta T \)

The formula for heat is \( q = mc\Delta T \), where \( c = 4.18\space J\cdot g^{-1}\cdot K^{-1} \) (and since a change in Kelvin is the same as a change in Celsius, we can use this for \( \Delta T \) in °C).
Rearranging for \( \Delta T \): \( \Delta T=\frac{q}{mc} \)
Substitute the values: \( \Delta T=\frac{70900\space J}{5500\space g\times4.18\space J\cdot g^{-1}\cdot K^{-1}} \)
First, calculate the denominator: \( 5500\times4.18 = 22990 \)
Then, \( \Delta T=\frac{70900}{22990}\approx3.08\space K \) (or °C)

Step3: Find the new temperature

Initial temperature \( T_i = 37.7\space ^\circ C \)
New temperature \( T_f = T_i+\Delta T \)
\( T_f = 37.7 + 3.08 = 40.78\space ^\circ C \)
Considering significant digits, the given values have 3 significant digits (5.50, 37.7, 70.9), so the answer should have 3 significant digits. Rounding 40.78 to three significant digits gives 40.8 °C. Wait, wait, let's recalculate \( \Delta T \) more accurately.
\( \Delta T=\frac{70900}{5500\times4.18}=\frac{70900}{22990}\approx3.084 \)
Then \( T_f = 37.7 + 3.084 = 40.784 \), which rounds to 40.8 °C? Wait, no, let's check the calculation again. Wait, 5500*4.18: 5500*4 = 22000, 5500*0.18 = 990, so total 22000 + 990 = 22990. Then 70900 / 22990: 70900 ÷ 22990 ≈ 3.084. Then 37.7 + 3.084 = 40.784, which is 40.8 °C when rounded to three significant digits? Wait, 37.7 has three, 5.50 has three, 70.9 has three. So the result should have three. Wait, but let's do the division more accurately. 70900 ÷ 22990: let's divide numerator and denominator by 10: 7090 ÷ 2299 ≈ 3.084. So 37.7 + 3.084 = 40.784, which is 40.8 °C? Wait, no, 3.084 added to 37.7: 37.7 + 3.084 = 40.784, which is 40.8 when rounded to three significant figures? Wait, 40.784: the first three significant digits are 4, 0, 7? No, wait, 40.784: the number is 40.784. The first significant digit is 4, second 0? No, wait, 40.784: significant digits are 4, 0, 7, 8, 4? No, no, 40.784: the leading zero after 4 is not significant? Wait, no, 40.784: the digits are 4 (1st), 0 (2nd, but it's between two non-zero digits, so it is significant), 7 (3rd), 8 (4th), 4 (5th). Wait, no, 40.784: the number is 4.0784 × 10^1, so the significant digits are 4, 0, 7, 8, 4. Wait, but our inputs: 5.50 (three sig figs), 37.7 (three), 70.9 (three). So when multiplying/dividing, the result should have three sig figs. So \( \Delta T \) was calculated as approximately 3.08 (three sig figs), then adding to 37.7 (three sig figs). When adding, the number of decimal places matters. 37.7 has one decimal place, 3.08 has two. So we go by the least number of decimal places, which is one. Wait, no, 37.7 is to the tenths place, 3.08 is to the hundredths. So when adding, 37.7 + 3.08 = 40.78, which should be rounded to the tenths place? Wait, no, 37.7 has one decimal place, so the sum should have one decimal place? Wait, no, 37.7 is 37.70 when considering two decimal places? No, 37.7 is precise to the tenths place. 3.08 is precise to the hundredths. So the sum is 40.78, which should be rounded to the tenths place: 40.8? Wait, no, 37.7 + 3.08 = 40.78. The rule for addition is that the result has the same number of decimal places as the term with the least number of decimal places. 37.7 has one decimal place, 3.08 has two. So we round to one decimal place: 40.8? Wait, 40.78 rounded to one decimal place is 40.8. Yes. But let's check the calculation of \( \Delta T \) again. Let's do 70900 / (5500*4.18):

5500 * 4.18 = 5500 * 4 + 5500 * 0.18 = 22000 + 990 = 22990

70900 / 22990 = let's divide numerator and denominator by 10: 7090 / 2299 ≈ 3.084

So 3.084, which is approximately 3.08 (three sig figs). Then 37.7 + 3.084 = 40.784, which is 40.8 when rounded to three significant figures? Wait, 40.784: the first three significant digits are 4, 0, 7? No, 40.784: the digits are 4 (1), 0 (2), 7 (3), 8 (4), 4 (5). Wait, no, 40.784 is 4.0784 × 10^1, so the significant digits are 4, 0, 7, 8, 4. So when we have three significant digits, it's 4.08 × 10^1, which is 40.8. Yes. So the new temperature is approximately 40.8 °C. Wait, but let's check with exact calculation:

\( q = mc\Delta T \)

\( \Delta T = \frac{q}{mc} = \frac{70900\space J}{5500\space g\times4.18\space J/g/K} \)

Calculate denominator: 5500 * 4.18 = 22990

70900 / 22990 ≈ 3.084 K (or °C)

Initial temperature: 37.7 °C

New temperature: 37.7 + 3.084 = 40.784 °C ≈ 40.8 °C (three significant figures)

Wait, but let's check the calculation again. Maybe I made a mistake in the division. 70900 divided by 22990:

22990 * 3 = 68970

70900 - 68970 = 1930

1930 / 22990 ≈ 0.084

So total \( \Delta T \approx 3.084 \), so adding to 37.7 gives 40.784, which is 40.8 when rounded to three significant figures.

Answer:

\( 40.8\space ^\circ C \)