Question

a copper rod that has a mass of 200.0 g has an initial temperature of 20.0°c and is heated to 40.0°c. if 1,540 j of heat are needed to heat the rod, what is the specific heat of copper? use $q = mc_pdelta t$. \bigcirc 0.0130 j/(g,°c) \bigcirc 0.0649 j/(g,°c) \bigcirc 0.193 j/(g,°c) \bigcirc 0.385 j/(g,°c)

Explanation:

Step1: Identify known values

We know the formula for heat transfer is \( q = mC_p\Delta T \), where \( q = 1540 \, \text{J} \), \( m = 200.0 \, \text{g} \), initial temperature \( T_i = 20.0^\circ\text{C} \), final temperature \( T_f = 40.0^\circ\text{C} \). First, calculate \( \Delta T \).
\( \Delta T = T_f - T_i = 40.0^\circ\text{C} - 20.0^\circ\text{C} = 20.0^\circ\text{C} \)

Step2: Rearrange formula to solve for \( C_p \)

From \( q = mC_p\Delta T \), we can rearrange to \( C_p = \frac{q}{m\Delta T} \)

Step3: Substitute values into the formula

Substitute \( q = 1540 \, \text{J} \), \( m = 200.0 \, \text{g} \), and \( \Delta T = 20.0^\circ\text{C} \) into the formula:
\( C_p = \frac{1540 \, \text{J}}{200.0 \, \text{g} \times 20.0^\circ\text{C}} \)
First, calculate the denominator: \( 200.0 \, \text{g} \times 20.0^\circ\text{C} = 4000 \, \text{g}^\circ\text{C} \)
Then, calculate \( C_p \): \( C_p = \frac{1540 \, \text{J}}{4000 \, \text{g}^\circ\text{C}} = 0.385 \, \text{J/(g}^\circ\text{C}) \)? Wait, no, wait, let's recalculate. Wait, 1540 divided by (20020) is 1540 / 4000 = 0.385? Wait, but that's not matching. Wait, no, maybe I made a mistake. Wait, 20020 is 4000. 1540/4000 = 0.385? But wait, the options have 0.385, but let's check again. Wait, no, wait the formula is \( q = mC\Delta T \), so \( C = q/(m\Delta T) \). So q is 1540 J, m is 200 g, delta T is 20 C. So 1540 / (20020) = 1540 / 4000 = 0.385? But wait, the correct specific heat of copper is 0.385 J/(g°C), but wait, let's check the calculation again. Wait, 20020 is 4000. 1540 divided by 4000 is 0.385. Wait, but let's check the options. The last option is 0.385 J/(g°C). Wait, but maybe I miscalculated. Wait, no, 20020 is 4000. 1540/4000 = 0.385. So the answer should be 0.385 J/(g°C). Wait, but let's check again. Wait, 1540 divided by (20020) = 1540/4000 = 0.385. Yes.

Wait, but wait, maybe I messed up the delta T. Wait, 40 - 20 is 20, that's correct. Mass is 200 g, correct. Heat is 1540 J. So 1540 / (200*20) = 0.385. So the answer is 0.385 J/(g°C), which is the last option.

Wait, but let me check again. 200 grams times 20 degrees is 4000 gram-degrees. 1540 joules divided by 4000 gram-degrees is 0.385 joules per gram-degree. So that's correct.

Answer:

0.385 J/(g,°C) (the last option, D. 0.385 J/(g,°C) assuming the options are labeled as A, B, C, D with D being 0.385 J/(g,°C))