Question

a 5.02 kg piece of copper metal is heated from 18.5 °c to 312.3 °c. calculate the heat absorbed (in kj) by the metal. be sure your answer has the correct number of significant digits. note: reference the phase change properties of pure substances table for additional information.

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 capacity, and \(\Delta T\) is the change in temperature. For copper, the specific heat capacity \(c = 0.385\ \text{J/g}^\circ\text{C}\) (from reference tables). First, convert the mass from kg to g: \(m = 5.02\ \text{kg} = 5020\ \text{g}\).

Step2: Calculate \(\Delta T\)

\(\Delta T=T_2 - T_1=312.3^\circ\text{C}- 18.5^\circ\text{C}=293.8^\circ\text{C}\).

Step3: Calculate heat in joules

Substitute into the formula: \(q=(5020\ \text{g})\times(0.385\ \text{J/g}^\circ\text{C})\times(293.8^\circ\text{C})\). First, calculate \(5020\times0.385 = 1932.7\), then \(1932.7\times293.8\approx1932.7\times294 = 1932.7\times(300 - 6)=1932.7\times300-1932.7\times6 = 579810 - 11596.2 = 568213.8\ \text{J}\).

Step4: Convert to kilojoules

Since \(1\ \text{kJ}=1000\ \text{J}\), divide by 1000: \(q=\frac{568213.8\ \text{J}}{1000}=568.2138\ \text{kJ}\). Now, consider significant digits. The mass has 3 significant digits, \(\Delta T\) has 4 (but when subtracting, the decimal places matter; \(312.3 - 18.5 = 293.8\), which has 4 significant digits in the decimal - free part? Wait, actually, the number of significant digits for the result should be based on the least number of significant digits in the multiplication/division. The mass (5.02) has 3, specific heat (0.385) has 3, \(\Delta T\) (293.8) has 4. So we take 3 significant digits. Rounding 568.2138 to 3 significant digits gives 568 kJ? Wait, no: 5.02 (3), 0.385 (3), 293.8 (4). So the product should have 3 significant digits. Let's recalculate more accurately:

\(q = 5020\times0.385\times293.8\)

\(5020\times0.385 = 5020\times\frac{385}{1000}=\frac{5020\times385}{1000}=\frac{1932700}{1000}=1932.7\)

\(1932.7\times293.8 = 1932.7\times(293 + 0.8)=1932.7\times293+1932.7\times0.8\)

\(1932.7\times293 = 1932.7\times(300 - 7)=1932.7\times300-1932.7\times7 = 579810 - 13528.9 = 566281.1\)

\(1932.7\times0.8 = 1546.16\)

So total \(q = 566281.1+1546.16 = 567827.26\ \text{J}=567.82726\ \text{kJ}\). Rounding to 3 significant digits: 568 kJ? Wait, no, 5.02 has 3, 0.385 has 3, so the result should have 3 significant digits. 567.82726 rounded to 3 significant digits is 568 kJ? Wait, 567.82726, the first three significant digits are 5, 6, 7, the next digit is 8, which is more than 5, so we round up the third digit: 568? Wait, no, 567.82726, the number is 567.8..., so when rounding to 3 significant digits, it's 568? Wait, 5.02 (3), 0.385 (3), so the answer should have 3 significant digits. So 568 kJ? Wait, let's check with a calculator:

\(5020\times0.385 = 1932.7\)

\(1932.7\times293.8 = 1932.7\times293.8 = 567827.26\ \text{J}=567.82726\ \text{kJ}\approx568\ \text{kJ}\) (with 3 significant digits).

Wait, maybe I made a mistake in the specific heat. Wait, the specific heat of copper is \(0.385\ \text{J/g}^\circ\text{C}\) or \(0.385\ \text{J/gK}\) (since the change in Celsius is the same as change in Kelvin). Let's confirm the formula again. The heat formula is \(q = mc\Delta T\), where \(\Delta T\) is the change in temperature. So all steps seem correct.

Answer:

\(568\) (or more accurately, if we consider the exact calculation: let's do it with more precision. \(5.02\ \text{kg}=5020\ \text{g}\), \(c = 0.385\ \text{J/g}^\circ\text{C}\), \(\Delta T=312.3 - 18.5 = 293.8^\circ\text{C}\).

\(q=5020\times0.385\times293.8\)

\(5020\times0.385 = 1932.7\)

\(1932.7\times293.8 = 1932.7\times293.8 = 567827.26\ \text{J}=567.82726\ \text{kJ}\). Rounding to three significant figures (since 5.02 has three, 0.385 has three), we get \(568\ \text{kJ}\). So the heat absorbed is \(\boxed{568}\) kJ (or if we consider that maybe the specific heat was taken as \(0.387\ \text{J/g}^\circ\text{C}\) (another common value), let's check:

If \(c = 0.387\ \text{J/g}^\circ\text{C}\):

\(q = 5020\times0.387\times293.8\)

\(5020\times0.387 = 5020\times\frac{387}{1000}=\frac{5020\times387}{1000}=\frac{1942740}{1000}=1942.74\)

\(1942.74\times293.8 = 1942.74\times(293 + 0.8)=1942.74\times293+1942.74\times0.8\)

\(1942.74\times293 = 1942.74\times(300 - 7)=1942.74\times300-1942.74\times7 = 582822 - 13599.18 = 569222.82\)

\(1942.74\times0.8 = 1554.192\)

\(q = 569222.82+1554.192 = 570777.012\ \text{J}=570.777012\ \text{kJ}\), which would round to \(571\ \text{kJ}\). But the standard specific heat of copper is \(0.385\ \text{J/g}^\circ\text{C}\) (at 25°C, for solid copper). So with \(c = 0.385\), we get \(568\ \text{kJ}\). So the answer is \(\boxed{568}\) kJ.