Math Problem Statement

Suppose that 0.48g of water at 25°C condenses on the surface of a 55g block of aluminum that is initially at 25°C. If the heat released during condensation goes only toward heating the metal, what is the final temperature (in °C) of the metal block? The specific heat capacity of aluminum is 0.9031/g °C.

Solution

To solve this problem, we need to find the final temperature of the aluminum block after absorbing the heat released during the condensation of water.

Step 1: Heat released by the condensation of water

The amount of heat released during condensation can be calculated using the latent heat of vaporization of water, which is approximately 2260J/g2260 \, \text{J/g}.

Given:

  • Mass of water, mwater=0.48gm_{\text{water}} = 0.48 \, \text{g}
  • Latent heat of vaporization, L=2260J/gL = 2260 \, \text{J/g}

The heat released by condensation is: Q=mwater×L=0.48g×2260J/g=1084.8JQ = m_{\text{water}} \times L = 0.48 \, \text{g} \times 2260 \, \text{J/g} = 1084.8 \, \text{J}

Step 2: Heat absorbed by the aluminum block

The heat absorbed by the aluminum block is used to raise its temperature. The relationship between heat, mass, specific heat capacity, and temperature change is given by: Q=maluminum×caluminum×ΔTQ = m_{\text{aluminum}} \times c_{\text{aluminum}} \times \Delta T where:

  • QQ is the heat absorbed by aluminum (1084.8 J),
  • maluminum=55gm_{\text{aluminum}} = 55 \, \text{g},
  • caluminum=0.903J/g°Cc_{\text{aluminum}} = 0.903 \, \text{J/g°C},
  • ΔT\Delta T is the change in temperature of the aluminum block.

Rearranging to solve for ΔT\Delta T: ΔT=Qmaluminum×caluminum=1084.8J55g×0.903J/g°C\Delta T = \frac{Q}{m_{\text{aluminum}} \times c_{\text{aluminum}}} = \frac{1084.8 \, \text{J}}{55 \, \text{g} \times 0.903 \, \text{J/g°C}} ΔT=1084.849.66521.84C\Delta T = \frac{1084.8}{49.665} \approx 21.84^\circ \text{C}

Step 3: Final temperature of the aluminum block

Since the initial temperature of the aluminum block is 25C25^\circ \text{C}, the final temperature will be: Tfinal=25C+ΔT=25C+21.84C=46.84CT_{\text{final}} = 25^\circ \text{C} + \Delta T = 25^\circ \text{C} + 21.84^\circ \text{C} = 46.84^\circ \text{C}

Thus, the final temperature of the aluminum block is approximately 46.84C\boxed{46.84^\circ \text{C}}.

Would you like more details or have any further questions?

Here are 5 related questions you could explore:

  1. What is the heat capacity of the aluminum block?
  2. How would the final temperature change if the block were copper instead of aluminum?
  3. How much heat is needed to raise 55 g of aluminum by 1°C?
  4. What happens if the mass of water condensed increases to 1 g?
  5. How does the specific heat of a material affect the rate of temperature change?

Tip: The larger the specific heat capacity of a material, the more heat is needed to raise its temperature.

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Math Problem Analysis

Mathematical Concepts

Thermodynamics
Heat Transfer
Specific Heat

Formulas

Q = m * L (Heat released by condensation)
Q = m * c * ΔT (Heat absorbed by the aluminum block)
ΔT = Q / (m * c) (Change in temperature)

Theorems

Law of Conservation of Energy

Suitable Grade Level

Grades 10-12