A 5.00 gram sample of aluminum pellets (specific heat capacity = 0.89 J/g*C) and a 10.00 gram sample of iron pellets (specific heat capacity = 0.45 J/g*C) are heated to 100.0 *C. The mixture of hot iron and aluminum is then dropped into 97.3 gram of water at 22.0*C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.
heat lost by Al + heat lost by Fe + heat gained by water = 0.
[mass Al x specific heat Al x (Tfinal-Tinitial)] + [mass Fe x specific heat Fe x (Tfinal-Tinitial)] + [mass water x speicific heat water x (Tfinal-Tinitial)] = 0
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To solve this problem, we can apply the principle of conservation of heat. The heat lost by the metal pellets will be gained by the water. The equation we can use is:
Q_lost = Q_gained
1. Calculate the heat lost by the metal pellets:
Q_lost = (mass of aluminum) x (specific heat capacity of aluminum) x (change in temperature)
+ (mass of iron) x (specific heat capacity of iron) x (change in temperature)
Change in temperature = final temperature - initial temperature
Given:
Mass of aluminum (m₁) = 5.00 g
Specific heat capacity of aluminum (c₁) = 0.89 J/g°C
Initial temperature of aluminum (T₁) = 100.0 °C
Mass of iron (m₂) = 10.00 g
Specific heat capacity of iron (c₂) = 0.45 J/g°C
Initial temperature of iron (T₂) = 100.0 °C
Q_lost = (5.00 g) x (0.89 J/g°C) x (final temperature - 100.0 °C)
+ (10.00 g) x (0.45 J/g°C) x (final temperature - 100.0 °C)
2. Calculate the heat gained by the water:
Q_gained = (mass of water) x (specific heat capacity of water) x (change in temperature)
Change in temperature = final temperature - initial temperature
Given:
Mass of water (m₃) = 97.3 g
Specific heat capacity of water (c₃) = 4.18 J/g°C
Initial temperature of water (T₃) = 22.0 °C
Q_gained = (97.3 g) x (4.18 J/g°C) x (final temperature - 22.0 °C)
3. Set Q_lost equal to Q_gained and solve for the final temperature:
(5.00 g) x (0.89 J/g°C) x (final temperature - 100.0 °C)
+ (10.00 g) x (0.45 J/g°C) x (final temperature - 100.0 °C)
= (97.3 g) x (4.18 J/g°C) x (final temperature - 22.0 °C)
Simplify and solve for the final temperature.
To calculate the final temperature of the metal and water mixture, we can use the principle of conservation of energy. The heat lost by the metal pellets will be equal to the heat gained by the water.
First, let's calculate the heat lost by the metal pellets:
Q1 = mcΔT
= (5.00 g)(0.89 J/g°C)(100.0°C - Tf)
= 445 J - 4.45Tf
Where Tf represents the final temperature of the metal.
Next, let's calculate the heat gained by the water:
Q2 = mcΔT
= (97.3 g)(4.18 J/g°C)(Tf - 22.0°C)
= 408.994Tf - 9036.3074
Where Tf represents the final temperature of the water.
Since there is no heat loss to the surroundings, the heat lost by the metal pellets is equal to the heat gained by the water:
445 J - 4.45Tf = 408.994Tf - 9036.3074
Simplifying the equation:
413.444Tf = 9481.3074
Tf = 22.95°C
Therefore, the final temperature of the metal and water mixture is approximately 22.95°C.