Math Problem Statement

A 19.4-gg piece of metal at 88.5 ∘C∘C is dropped into 41 gg of water at 23.5 ∘C∘C. All the heat lost by the metal is used to heat the water, and the final temperature of the water and metal is 28.0 ∘C∘C. Calculate the specific heat, in J/g ∘CJ/g ∘C, of the metal.

Express your answer to two significant figures.

Solution

To calculate the specific heat (cc) of the metal, we'll again use the principle of conservation of energy, where the heat lost by the metal is equal to the heat gained by the water.

Given:

  • Mass of metal (mmetalm_{\text{metal}}) = 19.4 g
  • Initial temperature of metal (Tinitial, metalT_{\text{initial, metal}}) = 88.5 °C
  • Final temperature (TfinalT_{\text{final}}) = 28.0 °C
  • Mass of water (mwaterm_{\text{water}}) = 41 g
  • Initial temperature of water (Tinitial, waterT_{\text{initial, water}}) = 23.5 °C
  • Specific heat of water (cwaterc_{\text{water}}) = 4.18 J/g°C

Steps:

  1. Calculate the temperature changes: ΔTmetal=TfinalTinitial, metal=28.088.5=60.5°C\Delta T_{\text{metal}} = T_{\text{final}} - T_{\text{initial, metal}} = 28.0 - 88.5 = -60.5 \, \text{°C} ΔTwater=TfinalTinitial, water=28.023.5=4.5°C\Delta T_{\text{water}} = T_{\text{final}} - T_{\text{initial, water}} = 28.0 - 23.5 = 4.5 \, \text{°C}

  2. Calculate the heat gained by the water: qwater=mwatercwaterΔTwaterq_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} qwater=41g4.18J/g°C4.5°Cq_{\text{water}} = 41 \, \text{g} \cdot 4.18 \, \text{J/g°C} \cdot 4.5 \, \text{°C} qwater=770.01Jq_{\text{water}} = 770.01 \, \text{J}

    The heat lost by the metal (qmetalq_{\text{metal}}) equals the heat gained by the water: qmetal=qwater=770.01Jq_{\text{metal}} = -q_{\text{water}} = -770.01 \, \text{J}

  3. Calculate the specific heat of the metal: qmetal=mmetalcmetalΔTmetalq_{\text{metal}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot \Delta T_{\text{metal}} 770.01J=19.4gcmetal(60.5°C)-770.01 \, \text{J} = 19.4 \, \text{g} \cdot c_{\text{metal}} \cdot (-60.5 \, \text{°C})

    Simplify to solve for cmetalc_{\text{metal}}: cmetal=770.01J19.4g(60.5°C)c_{\text{metal}} = \frac{-770.01 \, \text{J}}{19.4 \, \text{g} \cdot (-60.5 \, \text{°C})} cmetal770.01J1173.7g°C0.656J/g°Cc_{\text{metal}} \approx \frac{770.01 \, \text{J}}{1173.7 \, \text{g°C}} \approx 0.656 \, \text{J/g°C}

    Rounded to two significant figures: cmetal0.66J/g°Cc_{\text{metal}} \approx 0.66 \, \text{J/g°C}

The specific heat of the metal is approximately 0.66 J/g°C.

Would you like more details, or do you have any questions?

Here are five related questions:

  1. How does the specific heat of a substance relate to its atomic or molecular structure?
  2. What is the significance of a substance having a high specific heat compared to a low one?
  3. How would the result change if the water had a different initial temperature?
  4. Can you calculate the amount of heat required to raise the temperature of the metal by a certain number of degrees?
  5. How would you experimentally determine the specific heat of a substance in a laboratory?

Tip: Always ensure you use the correct number of significant figures in your final answer, matching the precision of the given data.

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Math Problem Analysis

Mathematical Concepts

Thermal Physics
Specific Heat

Formulas

q = m \cdot c \cdot \Delta T

Theorems

Conservation of Energy

Suitable Grade Level

High School