If a 0.55 g piece of copper $( Cp =0.387 J / gC )$ at 100 C is placed in 50 g of water $( Cp =4.18$ $J / gC$ ) initially at 23 C , what would the final temperature of the mixture be?
Answer (2)
The heat lost by the copper is equal to the heat gained by the water.
Calculate the heat lost by the copper: Q C u = m C u × C p C u × ( T C u , i − T f ) .
Calculate the heat gained by the water: Q H 2 O = m H 2 O × C p H 2 O × ( T f − T H 2 O , i ) .
Solve for the final temperature T f : T f = m C u × C p C u + m H 2 O × C p H 2 O m C u × C p C u × T C u , i + m H 2 O × C p H 2 O × T H 2 O , i ≈ 23.078 .
The final temperature of the mixture is 23.078 °C.
Explanation
Problem Analysis We are given a calorimetry problem where a piece of copper at a higher temperature is placed into water at a lower temperature. Our goal is to find the final temperature of the mixture once thermal equilibrium is reached. We will use the principle that the heat lost by the copper is equal to the heat gained by the water, assuming no heat is lost to the surroundings.
Heat Transfer Formula The heat lost or gained by a substance can be calculated using the formula: Q = m Cp Δ T , where:
Q is the heat transferred
m is the mass of the substance
Cp is the specific heat capacity of the substance
Δ T is the change in temperature
Defining Variables Let's define the variables:
m C u = 0.55 g (mass of copper)
C p C u = 0.387 J/g°C (specific heat of copper)
T C u , i = 100 °C (initial temperature of copper)
m H 2 O = 50 g (mass of water)
C p H 2 O = 4.18 J/g°C (specific heat of water)
T H 2 O , i = 23 °C (initial temperature of water)
T f = final temperature of the mixture (what we want to find)
Heat Lost by Copper The heat lost by the copper is: Q C u = m C u C p C u ( T C u , i − T f ) = 0.55 × 0.387 × ( 100 − T f )
Heat Gained by Water The heat gained by the water is: Q H 2 O = m H 2 O C p H 2 O ( T f − T H 2 O , i ) = 50 × 4.18 × ( T f − 23 )
Equating Heat Transfer Since the heat lost by the copper equals the heat gained by the water, we have: Q C u = Q H 2 O 0.55 × 0.387 × ( 100 − T f ) = 50 × 4.18 × ( T f − 23 )
Solving for Final Temperature Now, let's solve for T f :
0.21285 × ( 100 − T f ) = 209 × ( T f − 23 ) 21.285 − 0.21285 T f = 209 T f − 4807 209.21285 T f = 4828.285 T f = 209.21285 4828.285 ≈ 23.078
Final Answer Therefore, the final temperature of the mixture is approximately 23.078 °C.
Examples
Calorimetry problems are commonly used in chemistry and materials science to determine the specific heat capacity of new materials or to analyze the heat transfer efficiency of different insulation materials. For example, engineers might use calorimetry to test how well a new type of building insulation maintains a stable temperature inside a building, which helps in designing energy-efficient structures. Similarly, food scientists use calorimetry to measure the caloric content of food products, providing nutritional information to consumers.
The final temperature of the mixture of copper and water can be calculated using the principle of conservation of energy. By setting the heat lost by copper equal to the heat gained by water, we find the final temperature to be approximately 23.08 °C. This demonstrates the heat exchange between two substances until thermal equilibrium is reached.
;
