Calorimetry: Specific Heat of a Metal & Final Temperature of Water Mixtures

Thermal Equilibrium & Energy Transfer

  • Whenever two objects are placed in thermal contact they exchange energy through particle collisions until they reach the same final temperature.
  • At equilibrium the heat (energy) lost by the hotter object is numerically equal (but opposite in sign) to the heat gained by the colder object.
    • q{\text{lost}} + q{\text{gained}} = 0
    • Equivalent statement: q{\text{gained}} = -q{\text{lost}}
  • Assumptions used in introductory calorimetry problems:
    • No energy escapes to the surrounding air or calorimeter walls (the system is isolated).
    • Phase changes do not occur during the temperature change (all water stays liquid, metal stays solid).

Core Formulae & Symbols

  • Heat (thermal energy): q
  • Mass: m (in grams for these problems)
  • Specific heat capacity: c (J·g^{-1}$·°C^{-1})
  • Temperature change: \Delta T = T{\text{final}} - T{\text{initial}}
  • Fundamental relationship: q = m c \Delta T
    • A positive q means heat gained; negative q means heat lost.

Problem-Type 1 – Metal Dropped into Water

Given Data

  • Mass of metal, m_{\text{metal}} = 54\;\text{g}
  • Mass of water, m{\text{H}2\text{O}} = 200\;\text{g}
  • Initial temperature of water, T{\text{H}2\text{O},i} = 20.6\;^{\circ}\text{C}
  • Initial temperature of metal (after hot-water bath), T_{\text{metal},i} = 99.2\;^{\circ}\text{C}
  • Common final temperature after mixing, T_f = 23.4\;^{\circ}\text{C}
  • Specific heat of water, c{\text{H}2\text{O}} = 4.184\;\text{J·g}^{-1}\,^{\circ}\text{C}^{-1}
  • Unknown: specific heat of the metal, c_{\text{metal}}

Heat-Balance Set-up

  1. Water gains heat:
    q{\text{H}2\text{O}} = m{\text{H}2\text{O}}\,c{\text{H}2\text{O}}\,(Tf - T{\text{H}_2\text{O},i})
  2. Metal loses heat:
    q{\text{metal}} = m{\text{metal}}\,c{\text{metal}}\,(Tf - T_{\text{metal},i})
  3. Because energy is conserved:
    q{\text{H}2\text{O}} = -q_{\text{metal}}

Numerical Calculation

  • Heat gained by water:
    q{\text{H}2\text{O}} = 200\;(4.184)\,(23.4 - 20.6)
    = 200 \times 4.184 \times 2.8
    \approx 2\,340\;\text{J}
  • Solve for c{\text{metal}}: c{\text{metal}} = \frac{-q{\text{H}2\text{O}}}{m{\text{metal}}\,(Tf - T_{\text{metal},i})}
    = \frac{-2\,340}{54\,(23.4 - 99.2)}
    = \frac{-2\,340}{54\,(-75.8)}
    \approx 0.572\;\text{J·g}^{-1}\,^{\circ}\text{C}^{-1}

Interpretation

  • A specific heat near 0.57 J·g^{-1}$·°C^{-1} suggests the metal might be iron or steel (typical values \approx 0.45$–$0.50 J·g^{-1}$·°C^{-1}) but could also represent an alloy; experimental error, container heat capacity, or heat loss to air can slightly inflate the value.
  • Conceptual checkpoints:
    • Final temperature must lie between initial temps (it does).
    • Sign convention: the metal’s \Delta T is negative, giving a positive c_{\text{metal}}.

Problem-Type 2 – Mixing Two Water Samples (Honors / Extra-Credit)

Given Data

  • Hot water mass, m_h = 51.6\;\text{g}
  • Hot water initial temp, T_{h,i} = 42.6\;^{\circ}\text{C}
  • Cold water mass, m_c = 131.2\;\text{g}
  • Cold water initial temp, T_{c,i} = 18.2\;^{\circ}\text{C}
  • Specific heat (same for both samples): c = 4.184\;\text{J·g}^{-1}\,^{\circ}\text{C}^{-1}
  • Unknown: common final temperature, T_f = x

Energy-Balance Equation

  1. Cold water gains heat: qc = mc c (x - T_{c,i})
  2. Hot water loses heat: qh = mh c (x - T_{h,i})
  3. Conservation of energy: qc = -qh

Algebraic Solution

Because c is the same for both terms, it cancels:
mc (x - 18.2) + mh (x - 42.6) = 0
Insert numbers:
131.2(x - 18.2) + 51.6(x - 42.6) = 0
Expand and collect x:
131.2x - 131.2\times18.2 + 51.6x - 51.6\times42.6 = 0
(131.2 + 51.6)x = 131.2\times18.2 + 51.6\times42.6
182.8x = 4\,578.8
x = \frac{4\,578.8}{182.8} \approx 25.1\;^{\circ}\text{C}

Plausibility Check

  • Final temp (≈25 °C) lies between 18.2 °C and 42.6 °C.
  • Closer to the cold sample’s initial temp because the cold portion has greater mass (131 g vs. 52 g).

Significance & Exam Note

  • Instructor labels this water-mixing algebra problem as Honors / extra-credit; it will not appear on the standard section of the next test.

Connections, Tips, & Practical Notes

  • The metal-in-water experiment is a standard calorimetry lab used to identify unknown metals; expect a near-identical problem on the next test.
  • Always watch signs:
    • If \Delta T is negative (object cools), q$$ is negative.
    • Treat “heat lost” values with a leading minus or move them to the opposite side of the equation.
  • Dimensional focus: keep units consistent (grams & °C). Convert to kilograms or Kelvin only if the specific-heat constant is defined that way.
  • Safety / ethics: When doing the lab, handle hot metals with tongs and avoid spills that could scald or damage equipment.
  • Real-world analogy: adding ice cubes (cold water) to a drink (warm liquid) uses the same energy-balance ideas; engineers also use calorimetry in material testing, food energy measurement, and thermal-management system design.

Common Pitfalls & How to Avoid Them

  • Forgetting to set the same final temperature for all components in contact.
  • Dropping negative signs, producing an unphysical negative specific heat.
  • Ignoring the heat capacity of the calorimeter cup itself (more advanced courses add that term).
  • Rounding too early; keep extra digits until the final step, then round (here to 3 sig-figs).