Calorimetry and Specific Heat Calculations

Simple Rules of Heat and Calorimetry

  • What is Calorimetry?:

    • Calorimetry is a way to measure how much heat energy moves when something warms up or cools down.

    • Scientists do this inside a special insulated cup called a calorimeter (like a thermos) that keeps heat from escaping.

  • How Heat Moves:

    • Heat always flows on its own from warm things to cool things.

    • It keeps moving until both things are at the exact same temperature, which is called thermal equilibrium.

    • In a perfectly sealed container, all the heat lost by the warm object goes into the cool object:

Heat lost=Heat gained\text{Heat lost} = \text{Heat gained}

The Easy Heat Formula

  • To calculate how much heat moved, use this simple recipe:

Heat transferred=c×m×ΔT\text{Heat transferred} = c \times m \times \Delta T

  • cc = Specific heat capacity (how much energy it takes to warm up 1 g1\,\text{g} of a material by 1 ∘C1\,^\circ\text{C}).

  • mm = Mass (how heavy the sample is in grams, g\text{g}).

  • ΔT\Delta T = Temperature change (how many degrees hotter or colder it got).

Step-by-Step Example: Dropping Hot Metal in Cool Water

  • The Story:

    • A warm piece of copper metal (145 g145\,\text{g} at 100 ∘C100\,^\circ\text{C}) is dropped into cool water (250 g250\,\text{g} at 25 ∘C25\,^\circ\text{C}).

    • They settle at a final temperature of 28.8 ∘C28.8\,^\circ\text{C}.

  • Step 1: Find how much heat the water gained:

    • Water temperature increase:

ΔTwater=28.8 ∘C−25 ∘C=3.8 ∘C\Delta T_{\text{water}} = 28.8\,^\circ\text{C} - 25\,^\circ\text{C} = 3.8\,^\circ\text{C}

  • Heat gained by water:

Heat gained=4.184 J g−1 ∘C−1×250 g×3.8 ∘C=3974.8 J\text{Heat gained} = 4.184\,\text{J}\,\text{g}^{-1}\,^\circ\text{C}^{-1} \times 250\,\text{g} \times 3.8\,^\circ\text{C} = 3974.8\,\text{J}

  • Step 2: Find the specific heat of copper:

    • The heat lost by copper equals the heat gained by water (3974.8 J3974.8\,\text{J}).

    • Copper temperature drop:

ΔTcopper=100 ∘C−28.8 ∘C=71.2 ∘C\Delta T_{\text{copper}} = 100\,^\circ\text{C} - 28.8\,^\circ\text{C} = 71.2\,^\circ\text{C}

  • Solve for copper's specific heat (ccopperc_{\text{copper}}):

c<em>copper=Heat lostm</em>copper×ΔTcopperc<em>{\text{copper}} = \frac{\text{Heat lost}}{m</em>{\text{copper}} \times \Delta T_{\text{copper}}}

ccopper=3974.8 J145 g×71.2 ∘C=0.385 J g−1 ∘C−1c_{\text{copper}} = \frac{3974.8\,\text{J}}{145\,\text{g} \times 71.2\,^\circ\text{C}} = 0.385\,\text{J}\,\text{g}^{-1}\,^\circ\text{C}^{-1}

  • Step 3: Heat needed for a 100 g100\,\text{g} sample of copper:

    • To warm a 100 g100\,\text{g} piece of copper by 1 ∘C1\,^\circ\text{C}:

Heat capacity=0.385 J g−1 ∘C−1×100 g=38.5 J ∘C−1\text{Heat capacity} = 0.385\,\text{J}\,\text{g}^{-1}\,^\circ\text{C}^{-1} \times 100\,\text{g} = 38.5\,\text{J}\,^\circ\text{C}^{-1}