Question

consider a 555. g sample of water. how much energy is absorbed as the water is heated from 6.0 °c to 92.0 °c? the specific heat of water is 1.00 cal/(g·°c). be sure each of your answer entries has the correct number of significant figures. note: reference the si prefixes table for additional information. part: 0 / 2 part 1 of 2 report the energy absorbed in calories.

Explanation:

Step1: Recall the heat formula

The formula for heat absorbed \( q \) is \( q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat, and \( \Delta T \) is the change in temperature.

Step2: Calculate the temperature change

\( \Delta T = T_2 - T_1 = 92.0^\circ\text{C} - 6.0^\circ\text{C} = 86.0^\circ\text{C} \)

Step3: Substitute values into the formula

Given \( m = 555\space g \), \( c = 1.00\space \frac{\text{cal}}{\text{g}\cdot^\circ\text{C}} \), and \( \Delta T = 86.0^\circ\text{C} \).
\( q = 555\space g \times 1.00\space \frac{\text{cal}}{\text{g}\cdot^\circ\text{C}} \times 86.0^\circ\text{C} \)
\( q = 555 \times 86.0 \space \text{cal} \)
\( q = 47730 \space \text{cal} \) (rounding to appropriate significant figures, since 555 has 3, 86.0 has 3, 1.00 has 3, so result should have 3 significant figures? Wait, 555 is 3, 86.0 is 3, 1.00 is 3. So 555*86.0 = 555*86 = let's calculate 555*80=44400, 555*6=3330, total 44400+3330=47730. Now, 555 is 3 sig figs, 86.0 is 3, so the product should be 3 sig figs? Wait, 555 has 3, 86.0 has 3, 1.00 has 3. So 555*86.0 = 47730, which can be written as \( 4.77 \times 10^4 \) cal? Wait, no, 555*86.0: 555*86 = 47730. Let's check the significant figures. The mass is 555 g (3 sig figs), specific heat 1.00 cal/g°C (3 sig figs), temperature change 86.0 °C (3 sig figs). So the multiplication: 3*3*3, so the result should have 3 sig figs. 47730 rounded to 3 sig figs is 47700? Wait, no, 47730: the first three digits are 4,7,7, the next digit is 3, which is less than 5, so it's 47700? Wait, no, 47730. Let's see, 555 is 3 sig figs, 86.0 is 3, so 555*86.0 = 555*86.0 = 47730. Now, 555 has 3, 86.0 has 3, so the product should have 3 sig figs. So 47730 rounded to 3 sig figs is 4.77×10⁴? Wait, 47730: the number is 47730, which is 4.773×10⁴. Rounding to 3 sig figs: 4.77×10⁴, which is 47700. Wait, but maybe the problem expects us to just calculate the exact value first. Wait, 555*86.0 = 555*86 = let's do 500*86=43000, 55*86=4730, so total 43000+4730=47730. So 47730 cal. But let's check the significant figures. The mass is 555 g (three significant figures), the temperature change is 92.0 - 6.0 = 86.0 °C (three significant figures, because 92.0 has three, 6.0 has two? Wait, 6.0 °C has two significant figures? Wait, 6.0: the decimal after 6 means the zero is significant, so 6.0 has two sig figs? Wait, no: 6.0 has two significant figures? Wait, 6.0 is two: the 6 and the 0. 92.0 is three: 9,2,0. So ΔT = 92.0 - 6.0 = 86.0? Wait, 92.0 - 6.0 = 86.0? Wait, 92.0 - 6.0 = 86.0? Let's calculate: 92.0 - 6.0 = 86.0? Yes, because 92.0 - 6.0 = 86.0 (the decimal is carried over). So ΔT is 86.0 °C (three significant figures, because 92.0 has three, 6.0 has two? Wait, no, when subtracting, the number of decimal places matters. 92.0 has one decimal place, 6.0 has one decimal place, so the result should have one decimal place? Wait, 92.0 - 6.0 = 86.0 (one decimal place). So ΔT is 86.0 °C (three significant figures, since 86.0 has three: 8,6,0). The mass is 555 g (three significant figures), specific heat is 1.00 cal/g°C (three significant figures). So when multiplying, the number of significant figures is determined by the least number? Wait, no: when multiplying/dividing, the result has the same number of significant figures as the least precise measurement. Here, all have three significant figures (mass: 3, specific heat: 3, ΔT: 3). So the result should have three significant figures. So 555 * 1.00 * 86.0 = 555 * 86.0 = 47730. Now, 47730 with three significant figures: the first three digits are 4,7,7, the next digit is 3, which is less than 5, so we keep the third digit as 7, so 47700? Wait, no, 47730: to three significant figures, it's 4.77×10⁴, which is 47700. But maybe the problem doesn't require strict sig fig rounding yet, just calculate the value. Let's do the calculation: 555 g * 1.00 cal/g°C * (92.0 - 6.0) °C. 92.0 - 6.0 = 86.0 °C. So 555 * 1.00 * 86.0 = 555 * 86.0. 555*80=44400, 555*6=3330, so 44400+3330=47730. So the heat absorbed is 47730 calories. But let's check the significant figures again. The mass is 555 (three sig figs), specific heat 1.00 (three), temperature change 86.0 (three). So the product is 47730, which can be written as 4.77×10⁴ cal (three sig figs) or 47700 cal (if we consider the trailing zero as not significant, but in scientific notation, 4.77×10⁴ is three sig figs). Alternatively, maybe the problem expects the exact value without rounding for now, so 47730 cal. Wait, but let's see: 555*86 = 47730. So the answer is 47730 cal? Or maybe 4.77×10⁴ cal? Wait, the problem says "Be sure each of your answer entries has the correct number of significant figures." Let's check the given values: 555 g (3 sig figs), 6.0 °C (2 sig figs? Wait, 6.0: the zero is significant, so 2 sig figs? Wait, 6.0 has two significant figures: 6 and 0. 92.0 has three: 9,2,0. So when subtracting, 92.0 - 6.0 = 86.0. The number of decimal places: 92.0 has one decimal place, 6.0 has one, so the result has one decimal place, so 86.0 (which has three significant figures, because the zero is after the decimal? No, 86.0 has three significant figures: 8,6,0. Because the zero is at the end of a number with a decimal, it's significant. So 86.0 has three sig figs. 555 has three, 1.00 has three. So all three measurements have three sig figs, so the product should have three. So 555*86.0 = 47730. Now, 47730: to three sig figs, we look at the fourth digit, which is 3, so we round down. So 47700? Wait, no, 47730: the first three digits are 4,7,7, the next digit is 3, so we keep the third digit as 7, so 47700. But 47700 has two significant figures? No, 47700: if we write it as 4.77×10⁴, that's three sig figs. So the correct answer with three significant figures is 4.77×10⁴ cal or 47700 cal (but 47700 could be ambiguous, so scientific notation is better). Wait, but maybe the problem doesn't care about sig figs for the calculation, just the formula. Let's proceed with the calculation:

\( q = mc\Delta T = 555\space g \times 1.00\space \frac{\text{cal}}{\text{g}\cdot^\circ\text{C}} \times (92.0 - 6.0)^\circ\text{C} \)

\( \Delta T = 86.0^\circ\text{C} \)

\( q = 555 \times 1.00 \times 86.0 = 47730\space \text{cal} \)

So the energy absorbed is 47730 calories (or 4.77×10⁴ cal with three significant figures).

Answer:

47730 (or 4.77×10⁴) cal