Math Problem Statement

Emily found a metallic solid in her backpack and determined its specific heat in J/g°C. Its mass was 80.77 g and she heated it to 100.0°C. It was placed in 100.0 g water at 25.0°C and the final temperature of the system reached 30.0°C. What was the solid?

Solution

To solve this problem, we need to use the principle of conservation of energy, specifically heat transfer between the metal and the water.

Step-by-Step Solution:

  1. Identify the Known Values:

    • Mass of the metal (mmetalm_{\text{metal}}): 80.77 g
    • Initial temperature of the metal (Tinitial, metalT_{\text{initial, metal}}): 100.0°C
    • Mass of water (mwaterm_{\text{water}}): 100.0 g
    • Specific heat of water (cwaterc_{\text{water}}): 4.18 J/g°C
    • Initial temperature of water (Tinitial, waterT_{\text{initial, water}}): 25.0°C
    • Final temperature of both water and metal (TfinalT_{\text{final}}): 30.0°C
  2. Calculate the Heat Gained by Water: The heat gained by the water (qwaterq_{\text{water}}) can be calculated using the formula: qwater=mwater×cwater×(TfinalTinitial, water)q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times (T_{\text{final}} - T_{\text{initial, water}}) Substituting the values: qwater=100.0g×4.18J/g°C×(30.025.0)°Cq_{\text{water}} = 100.0 \, \text{g} \times 4.18 \, \text{J/g°C} \times (30.0 - 25.0) \, \text{°C} qwater=100.0×4.18×5.0=2090Jq_{\text{water}} = 100.0 \times 4.18 \times 5.0 = 2090 \, \text{J}

  3. Apply the Conservation of Energy: Since the heat lost by the metal equals the heat gained by the water, we have: qmetal=qwater=2090Jq_{\text{metal}} = q_{\text{water}} = 2090 \, \text{J}

  4. Calculate the Specific Heat of the Metal: Using the formula for heat transfer, q=m×c×ΔTq = m \times c \times \Delta T, we rearrange to find the specific heat (cmetalc_{\text{metal}}) of the metal: cmetal=qmetalmmetal×(Tinitial, metalTfinal)c_{\text{metal}} = \frac{q_{\text{metal}}}{m_{\text{metal}} \times (T_{\text{initial, metal}} - T_{\text{final}})} Substituting the values: cmetal=2090J80.77g×(100.030.0)°Cc_{\text{metal}} = \frac{2090 \, \text{J}}{80.77 \, \text{g} \times (100.0 - 30.0) \, \text{°C}} cmetal=209080.77×700.4605J/g°Cc_{\text{metal}} = \frac{2090}{80.77 \times 70} \approx 0.4605 \, \text{J/g°C}

  5. Identify the Metal: From the given choices, the specific heat value closest to 0.4605 J/g°C corresponds to iron.

Answer:

The metallic solid is iron.

Would you like further explanation on any step?


Here are some follow-up questions to deepen understanding:

  1. What is the principle of conservation of energy and how is it applied in this scenario?
  2. How would the answer change if the mass of water was different?
  3. Why is it important to reach thermal equilibrium in this type of problem?
  4. How does specific heat relate to a material's ability to retain heat?
  5. Can you think of real-life applications of specific heat capacity in material selection?

Tip: Always check units for consistency when working with heat transfer problems to avoid calculation errors.

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Math Problem Analysis

Mathematical Concepts

Thermodynamics
Heat Transfer
Specific Heat Capacity

Formulas

q = mcΔT
q_water = m_water * c_water * (T_final - T_initial, water)
q_metal = m_metal * c_metal * (T_initial, metal - T_final)

Theorems

Conservation of Energy

Suitable Grade Level

Grades 9-12