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A $5.00-\mathrm{g}$ sample of aluminum pellets (specific heat capacity $=$ $0.89 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}$ ) and a $10.00-\mathrm{g}$ sample of iron pellets (specific heat capacity $=0.45 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}$ ) are heated to $100.0^{\circ} \mathrm{C}$. The mixture of hot iron and aluminum is then dropped into $97.3 \mathrm{~g}$ water at $22.0^{\circ} \mathrm{C}$. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

Name: Problem 55, Chapter 6: Chemistry
Uploaded: 2020-04-10T19:42:43-08:00
Duration: 2 min 49 s
Channel: Rebecca Wallace
Description: A 5.00-g sample of aluminum pellets (specific heat capacity = 0.89 J / ^∘ C ·g ) and a 10.00-g sample of iron pellets (specific heat capacity =0.45 J / ^∘ C ·g ) are heated to 100.0^∘ C. The mixture of hot iron and aluminum is then dropped into 97.3 g water at 22.0^∘ C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

   A $5.00-\mathrm{g}$ sample of aluminum pellets (specific heat capacity $=$ $0.89 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}$ ) and a $10.00-\mathrm{g}$ sample of iron pellets (specific heat capacity $=0.45 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}$ ) are heated to $100.0^{\circ} \mathrm{C}$. The mixture of hot iron and aluminum is then dropped into $97.3 \mathrm{~g}$ water at $22.0^{\circ} \mathrm{C}$. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

Chemistry

Steven S. Zumdahl,… 8th Edition

Chapter 6, Problem 55

Instant Answer

Solved by Expert Rebecca Wallace

04/10/2020

Step 1

First, we need to find the total heat gained by the water. To do this, we'll use the formula: q = mcΔT, where q is the heat gained or lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Show more…

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A $5.00-\mathrm{g}$ sample of aluminum pellets (specific heat capacity $=$ $0.89 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}$ ) and a $10.00-\mathrm{g}$ sample of iron pellets (specific heat capacity $=0.45 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}$ ) are heated to $100.0^{\circ} \mathrm{C}$. The mixture of hot iron and aluminum is then dropped into $97.3 \mathrm{~g}$ water at $22.0^{\circ} \mathrm{C}$. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

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Key Concepts

Specific Heat Capacity

This is a physical property that measures the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). It determines how substances absorb and release heat and is crucial for calculating temperature changes in calorimetry experiments.

Conservation of Energy

In thermodynamics, the principle of conservation of energy states that energy in an isolated system remains constant. In calorimetry, this concept implies that heat lost by warmer substances is equal to the heat gained by cooler substances, enabling the calculation of an equilibrium temperature.

Calorimetry

Calorimetry is the study of measuring the amount of heat exchanged in chemical reactions or physical changes. It employs the specific heat capacity values of substances and the conservation of energy to determine temperature changes in a system, assuming no heat loss to the environment.

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