What is the final tempurature when a 3.0 kg gold bar at 99C is dropped into 0.22 kg of water at 25C Specific heat of gold is 129 j/kg x C
The sum of the heats gained is zero (something loses heat).
Heatgainedwater+heatgainedgold=0
.22*Cw*(Tf-25)+3*Cg*(Tf-99)=0
solve for Tf
A*t =v-u at=v-u add u to both side at*u=v-at
To find the final temperature when the gold bar is dropped into water, we can use the principle of conservation of energy.
First, we need to calculate the amount of heat transferred from the gold bar to the water.
The amount of heat transferred can be calculated using the equation:
Q = m * c * ΔT
Where:
Q represents the heat transferred
m is the mass of the substance (gold or water)
c is the specific heat capacity of the substance
ΔT is the change in temperature
Let's calculate the heat transferred from the gold bar to the water:
- The mass of the gold bar is 3.0 kg
- The specific heat capacity of gold is 129 J/kg°C
- The initial temperature of the gold bar is 99°C
ΔT for the gold bar can be calculated as:
ΔT = final temperature - initial temperature
Let's assume the final temperature is T (in °C).
ΔT = T - 99°C
Now, let's calculate the heat transferred from the gold bar:
Qgold = m * c * ΔT
= (3.0 kg) * (129 J/kg°C) * (T - 99°C)
Next, let's calculate the heat transferred to the water:
- The mass of the water is 0.22 kg
- The specific heat capacity of water is 4186 J/kg°C
- The initial temperature of the water is 25°C
ΔT for the water can be calculated as:
ΔT = final temperature - initial temperature
ΔT = T - 25°C
Now, let's calculate the heat transferred to the water:
Qwater = m * c * ΔT
= (0.22 kg) * (4186 J/kg°C) * (T - 25°C)
According to the principle of conservation of energy, the heat transferred from the gold bar to the water (Qgold) should be equal to the heat transferred to the water (Qwater).
Therefore, we have the equation:
Qgold = Qwater
Substituting the previously calculated expressions for Qgold and Qwater:
(3.0 kg) * (129 J/kg°C) * (T - 99°C) = (0.22 kg) * (4186 J/kg°C) * (T - 25°C)
Now, we can solve this equation and find the value of T, which represents the final temperature.
1. Simplify the equation:
(3.0 kg) * (129 J/kg°C) * T - (3.0 kg) * (129 J/kg°C) * 99°C = (0.22 kg) * (4186 J/kg°C) * T - (0.22 kg) * (4186 J/kg°C) * 25°C
2. Rearrange the equation to isolate T terms on one side and non-T terms on the other side:
(3.0 kg) * (129 J/kg°C) * T - (0.22 kg) * (4186 J/kg°C) * T = (0.22 kg) * (4186 J/kg°C) * 25°C - (3.0 kg) * (129 J/kg°C) * 99°C
3. Calculate the values on each side:
(3.0 kg * 129 J/kg°C - 0.22 kg * 4186 J/kg°C) * T = (0.22 kg * 4186 J/kg°C * 25°C) - (3.0 kg * 129 J/kg°C * 99°C)
4. Solve for T:
T = [(0.22 kg * 4186 J/kg°C * 25°C) - (3.0 kg * 129 J/kg°C * 99°C)] / (3.0 kg * 129 J/kg°C - 0.22 kg * 4186 J/kg°C)
Evaluating this expression will give us the final temperature.