a 15.7 g aluminum block is warmed to 53.2 degrees celsius and plunged into an insulated beaker containing 32.5 g of water initially at 24.5 degrees celsius. the aluminum and the water are allowed to come to thermal equilibrium. assuming no heat is lost, what is the final temperature of water and aluminum?
heat lost by Al + heat gained by H2O = 0
[mass Al x specific heat Al x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Solve for Tfinal.
To calculate the final temperature of the water and aluminum, we can use the principle of conservation of energy. The heat lost by the aluminum block will be equal to the heat gained by the water.
First, let's calculate the heat lost by the aluminum block:
q_aluminum = m_aluminum * c_aluminum * ΔT_aluminum
Where:
m_aluminum = mass of the aluminum block
c_aluminum = specific heat capacity of aluminum (0.897 J/g°C)
ΔT_aluminum = change in temperature of the aluminum block
Given:
m_aluminum = 15.7 g
ΔT_aluminum = (final temperature - initial temperature) = (final temperature - 53.2°C)
Next, let's calculate the heat gained by the water:
q_water = m_water * c_water * ΔT_water
Where:
m_water = mass of the water
c_water = specific heat capacity of water (4.18 J/g°C)
ΔT_water = change in temperature of the water
Given:
m_water = 32.5 g
ΔT_water = (final temperature - initial temperature) = (final temperature - 24.5°C)
Since the system is insulated and no heat is lost, we can equate the two equations:
q_aluminum = q_water
m_aluminum * c_aluminum * ΔT_aluminum = m_water * c_water * ΔT_water
Substituting the given values:
15.7 g * 0.897 J/g°C * (final temperature - 53.2°C) = 32.5 g * 4.18 J/g°C * (final temperature - 24.5°C)
Simplifying the equation:
14.1283 * (final temperature - 53.2) = 134.35 * (final temperature - 24.5)
14.1283 * final temperature - 14.1283(53.2) = 134.35 * final temperature - 134.35(24.5)
14.1283 * final temperature - 752.7576 = 134.35 * final temperature - 3288.675
Rearranging the equation:
120.2217 * final temperature - 134.35 * final temperature = -3288.675 + 752.7576
-14.1283 * final temperature = -2535.9174
final temperature = -2535.9174 / -14.1283
final temperature ≈ 179.35°C
Therefore, the final temperature of the water and aluminum block is approximately 179.35°C.
To find the final temperature of the water and aluminum, we need to use the principle of conservation of energy, specifically the equation for heat transfer:
\( Q = mcΔT \)
where:
Q = heat transferred (in calories or joules)
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
First, let's determine the heat transferred from the aluminum block to the water.
1. Determine the specific heat capacity of aluminum:
The specific heat capacity of aluminum is 0.897 J/g°C or 0.214 cal/g°C.
2. Calculate the heat transferred from the aluminum block to the water:
\( Q_{\text{aluminum}} = m_{\text{aluminum}} \times c_{\text{aluminum}} \times \Delta T_{\text{aluminum}} \)
where:
\( m_{\text{aluminum}} = 15.7 \) g (mass of aluminum)
\( \Delta T_{\text{aluminum}} = T_{\text{final}} - T_{\text{initial}} = T_{\text{final}} - 53.2 \) °C (change in temperature)
Substituting the values we have:
\( Q_{\text{aluminum}} = 15.7 \times 0.897 \times (T_{\text{final}} - 53.2) \) J (or \( Q_{\text{aluminum}} = 15.7 \times 0.214 \times (T_{\text{final}} - 53.2) \) cal)
Next, let's determine the heat transferred from the water to the aluminum.
3. Determine the specific heat capacity of water:
The specific heat capacity of water is approximately 4.18 J/g°C or 1 cal/g°C.
4. Calculate the heat transferred from the water to the aluminum:
\( Q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T_{\text{water}} \)
where:
\( m_{\text{water}} = 32.5 \) g (mass of water)
\( \Delta T_{\text{water}} = T_{\text{final}} - T_{\text{initial}} = T_{\text{final}} - 24.5 \) °C (change in temperature)
Substituting the values we have:
\( Q_{\text{water}} = 32.5 \times 4.18 \times (T_{\text{final}} - 24.5) \) J (or \( Q_{\text{water}} = 32.5 \times 1 \times (T_{\text{final}} - 24.5) \) cal)
Since energy is conserved, the heat transferred from the aluminum to the water is equal to the heat transferred from the water to the aluminum:
\( Q_{\text{aluminum}} = Q_{\text{water}} \)
Now we can set these two equations equal to each other and solve for the final temperature \( T_{\text{final}} \):
\( 15.7 \times 0.897 \times (T_{\text{final}} - 53.2) = 32.5 \times 4.18 \times (T_{\text{final}} - 24.5) \)
Simplify and solve for \( T_{\text{final}} \), which will give you the final temperature of both the water and aluminum.