a 20.94-g sample of an unknown metal is heated to 99.4 degrees Celsius in a hot water bath until thermal equilibrium is reached. The metal is quickly transferred to 100 mL of water at 22.0 degrees Celsius contained in a styrofoam cup. The thermal equilibrium temperature of the metal plus water mixture is 24.6 degrees Celsius. What is the Specific heat capacity of the metal?
To find the specific heat capacity of the metal, we can use the heat transfer equation:
q = m * c * ΔT
Where:
q: heat transferred
m: mass of the metal
c: specific heat capacity of the metal
ΔT: temperature change
First, let's find the heat transferred from the metal to the water:
Heat transferred from the metal (q1) = Heat gained by the water (q2)
q1 = q2
Now, let's calculate q1:
q1 = m * c * ΔT1
Where:
m: mass of the metal (20.94 g)
c: specific heat capacity of the metal (unknown)
ΔT1: temperature change of the metal
Next, let's calculate q2:
q2 = m2 * c2 * ΔT2
Where:
m2: mass of the water (100 mL = 100 g)
c2: specific heat capacity of water (4.18 J/g°C)
ΔT2: temperature change of the water (24.6°C - 22.0°C)
Since q1 = q2, we can set up the equation:
m * c * ΔT1 = m2 * c2 * ΔT2
Substituting the given values, we get:
20.94 g * c * (99.4°C - 24.6°C) = 100 g * 4.18 J/g°C * (24.6°C - 22.0°C)
Now, let's solve for c:
(20.94 g * c * 74.8°C) = (100 g * 4.18 J/g°C * 2.6°C)
c = (100 g * 4.18 J/g°C * 2.6°C) / (20.94 g * 74.8°C)
c ≈ 1.77 J/g°C
Therefore, the specific heat capacity of the metal is approximately 1.77 J/g°C.
To find the specific heat capacity of the metal, we can use the equation:
\( q = mcΔT \)
where:
- \( q \) is the heat gained or lost by the substance (in this case, the metal and water mixture),
- \( m \) is the mass of the substance (in this case, the water),
- \( c \) is the specific heat capacity of the substance (in this case, the metal),
- \( ΔT \) is the change in temperature (final temperature - initial temperature).
Let's break down the information given in the problem:
1. The mass of the metal sample is 20.94 g.
2. The temperature rise of the metal sample is \( ΔT_m = T_{final} - T_{initial} = 24.6 °C - 99.4 °C \).
3. The mass of water is 100 mL. Since 1 mL of water has a mass of 1 g, the mass of water is 100 g.
4. The temperature rise of the water is \( ΔT_w = T_{final} - T_{initial} = 24.6 °C - 22.0 °C \).
First, let's calculate the heat absorbed by the water using the equation:
\( q_w = m_wc_wΔT_w \)
where:
- \( q_w \) is the heat gained by the water,
- \( m_w \) is the mass of the water,
- \( c_w \) is the specific heat capacity of water (which is approximately 4.18 J/g°C).
\( q_w = (100 \, \text{g}) \times (4.18 \, \text{J/g°C}) \times (24.6 °C - 22.0 °C) \)
Simplifying the calculation:
\( q_w = 100 \, \text{g} \times 4.18 \, \text{J/g°C} \times 2.6 °C \)
Now, let's calculate the heat lost by the metal using the equation:
\( q_m = m_mc_mΔT_m \)
where:
- \( q_m \) is the heat lost by the metal,
- \( m_m \) is the mass of the metal,
- \( c_m \) is the specific heat capacity of the metal (what we are trying to find).
\( q_m = (20.94 \, \text{g}) \times c_m \times (-74.8 °C) \)
Since the metal and water reach thermal equilibrium, the heat gained by the water is equal to the heat lost by the metal:
\( q_w = q_m \)
We have the values for \( q_w \) and \( q_m \), so we can equate them:
\( (100 \, \text{g}) \times (4.18 \, \text{J/g°C}) \times (24.6 °C - 22.0 °C) = (20.94 \, \text{g}) \times c_m \times (-74.8 °C) \)
Now, we can solve for \( c_m \):
\( c_m = \frac{{(100 \, \text{g}) \times (4.18 \, \text{J/g°C}) \times (24.6 °C - 22.0 °C)}}{{(20.94 \, \text{g}) \times (-74.8 °C)}} \)
Evaluating the expression:
\( c_m = \frac{{100 \, \text{g} \times 4.18 \, \text{J/g°C} \times 2.6 °C}}{{20.94 \, \text{g} \times 74.8 °C}} \)
Simplifying the calculation:
\( c_m \approx 0.889 \, \text{J/g°C} \)
Therefore, the specific heat capacity of the metal is approximately 0.889 J/g°C.
[mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial) = 0
Plug in all the numbers and solve for specific heat metal.
Post your work if you run into trouble.