Question

calculate the energy required to heat 481.0 mg of silver from -0.4 °c to 12.2 °c. assume the specific heat capacity of silver under these conditions is 0.235 j·g⁻¹·k⁻¹. round your answer to 3 significant digits.

Explanation:

Step1: Convert mass to grams

The mass of silver is \( 481.0 \, \text{mg} \). Since \( 1 \, \text{g} = 1000 \, \text{mg} \), we convert milligrams to grams: \( m = \frac{481.0 \, \text{mg}}{1000 \, \text{mg/g}} = 0.4810 \, \text{g} \).

Step2: Calculate temperature change

The initial temperature \( T_i = -0.4 \, ^\circ\text{C} \) and the final temperature \( T_f = 12.2 \, ^\circ\text{C} \). The temperature change \( \Delta T = T_f - T_i = 12.2 \, ^\circ\text{C} - (-0.4 \, ^\circ\text{C}) = 12.6 \, ^\circ\text{C} \). (Note: A change in Celsius is the same as a change in Kelvin, so \( \Delta T \) in K is also \( 12.6 \, \text{K} \).)

Step3: Use the heat formula \( q = mc\Delta T \)

The specific heat capacity \( c = 0.235 \, \text{J·g}^{-1}·\text{K}^{-1} \), mass \( m = 0.4810 \, \text{g} \), and \( \Delta T = 12.6 \, \text{K} \). Plugging in the values: \( q = 0.4810 \, \text{g} \times 0.235 \, \text{J·g}^{-1}·\text{K}^{-1} \times 12.6 \, \text{K} \).

First, multiply \( 0.4810 \times 0.235 \): \( 0.4810 \times 0.235 = 0.113035 \).

Then, multiply by \( 12.6 \): \( 0.113035 \times 12.6 = 1.424241 \, \text{J} \).

Step4: Round to 3 significant digits

The value \( 1.424241 \, \text{J} \) rounded to 3 significant digits is \( 1.42 \, \text{J} \) (wait, no, let's check the multiplication again. Wait, \( 0.4810 \times 0.235 = 0.4810 \times 0.235 \). Let's recalculate: \( 0.4810 \times 0.235 \):

\( 0.4810 \times 0.2 = 0.0962 \)

\( 0.4810 \times 0.03 = 0.01443 \)

\( 0.4810 \times 0.005 = 0.002405 \)

Adding them up: \( 0.0962 + 0.01443 + 0.002405 = 0.113035 \). Then \( 0.113035 \times 12.6 \):

\( 0.113035 \times 12 = 1.35642 \)

\( 0.113035 \times 0.6 = 0.067821 \)

Adding those: \( 1.35642 + 0.067821 = 1.424241 \). Now, rounding to 3 significant digits: the first three significant digits are 1, 4, 2, and the next digit is 4, which is less than 5, so we keep it as \( 1.42 \, \text{J} \)? Wait, no, wait: 0.4810 has 4 significant digits, 0.235 has 3, 12.6 has 3. When multiplying, the result should have 3 significant digits. So 1.424241 rounded to 3 significant digits is \( 1.42 \, \text{J} \)? Wait, no, 1.424241: the third significant digit is 2, the next digit is 4, so we round down, so 1.42. Wait, but let's check the calculation again. Wait, maybe I made a mistake in the temperature change? Wait, \( -0.4 \) to \( 12.2 \): \( 12.2 - (-0.4) = 12.6 \), that's correct. Mass: 481.0 mg is 0.4810 g, correct. Specific heat: 0.235 J/g·K. So \( q = 0.4810 \times 0.235 \times 12.6 \). Let's compute this with a calculator: \( 0.4810 \times 0.235 = 0.113035 \); \( 0.113035 \times 12.6 = 1.424241 \). Rounded to 3 significant digits: 1.42 J? Wait, no, 1.424241: the first three significant figures are 1, 4, 2, and the next digit is 4, so we round to 1.42. Wait, but maybe I messed up the significant digits. Wait, 481.0 mg is 4 significant digits, 0.235 is 3, 12.6 is 3. So the least number of significant digits in the multiplication is 3, so the answer should have 3. So 1.42 J. Wait, but let's check again. Wait, 0.4810 is 4 sig figs, 0.235 is 3, 12.6 is 3. So when multiplying, the result should have 3 sig figs. So 1.424241 rounded to 3 sig figs is 1.42. Wait, but maybe I made a mistake in the calculation. Wait, let's do the multiplication step by step:

\( 0.4810 \times 0.235 = 0.4810 \times 0.235 \)

\( 0.4810 \times 0.2 = 0.0962 \)

\( 0.4810 \times 0.03 = 0.01443 \)

\( 0.4810 \times 0.005 = 0.002405 \)

Sum: \( 0.0962 + 0.01443 = 0.11063 + 0.002405 = 0.113035 \)

Then \( 0.113035 \times 12.6 \):

\( 0.113035 \times 10 = 1.13035 \)

\( 0.113035 \times 2 = 0.22607 \)

\( 0.113035 \times 0.6 = 0.067821 \)

Sum: \( 1.13035 + 0.22607 = 1.35642 + 0.067821 = 1.424241 \)

Yes, that's correct. So rounding to 3 significant digits: 1.42 J. Wait, but let me check with a calculator: 0.481 * 0.235 * 12.6. Let's compute 0.481 * 0.235 first: 0.481 * 0.235 = 0.113035. Then 0.113035 * 12.6 = 1.424241. So 1.42 J when rounded to 3 significant digits.

Answer:

\( 1.42 \, \text{J} \)