A piece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °C and dropped into 84 g of water (with an initial temperature of 20 °C) in a calorimeter. The final temperature of the system is 52.1°C. The specific heat of water is 4.184 J/g*⁰C. What is the specific heat of the metal?
(a)0.171
(b)0.343
(c)1.717
(d)3.433
To find the specific heat of the metal, we can use the principle of conservation of energy. The heat lost by the metal will be gained by the water.
The heat gained by the water can be calculated using the formula:
q_water = m_water * c_water * ΔT_water
where:
q_water = heat gained by water
m_water = mass of water
c_water = specific heat of water
ΔT_water = change in temperature of water
Given:
m_water = 84 g
c_water = 4.184 J/g*⁰C
ΔT_water = (final temperature of system - initial temperature of water)
= 52.1 °C - 20 °C
= 32.1 °C
q_water = 84 g * 4.184 J/g*⁰C * 32.1 °C
= 11121.1344 J
The heat lost by the metal is equal to the heat gained by the water:
q_metal = q_water
The heat lost by the metal can be calculated using the formula:
q_metal = m_metal * c_metal * ΔT_metal
where:
q_metal = heat lost by metal
m_metal = mass of metal
c_metal = specific heat of metal
ΔT_metal = change in temperature of metal
Given:
m_metal = 68.6 g
ΔT_metal = (final temperature of system - initial temperature of metal)
= 52.1 °C - 100 °C
= -47.9 °C (negative because the metal is losing heat)
q_metal = 68.6 g * c_metal * (-47.9 °C)
Since q_metal = q_water, we can set the equations equal to each other:
68.6 g * c_metal * (-47.9 °C) = 11121.1344 J
To solve for c_metal, we can rearrange the equation:
c_metal = 11121.1344 J / (68.6 g * (-47.9 °C))
Calculating the specific heat of the metal:
c_metal = -0.3428 J/g*⁰C
Since specific heat cannot be negative, we can take the absolute value:
c_metal = 0.3428 J/g*⁰C
Therefore, the specific heat of the metal is approximately 0.343 J/g*⁰C.
The correct option is (b) 0.343.
To find the specific heat of the metal, we can use the principle of conservation of energy. When the metal is dropped into the water, heat flows from the metal to the water until they reach thermal equilibrium. We can express this using the equation:
Heat gained by water = Heat lost by metal
The heat gained by the water can be calculated using the equation:
qwater = m * cw * ΔT
where m is the mass of water, cw is the specific heat of water, and ΔT is the change in temperature of the water (final temperature - initial temperature).
The heat lost by the metal can be calculated using the equation:
qmetal = m * cm * ΔT
where m is the mass of the metal, cm is the specific heat of the metal, and ΔT is the change in temperature of the metal (final temperature - initial temperature).
Since the heat lost by the metal is equal to the heat gained by the water, we can set up the equation:
mwater * cw * ΔTwater = mmetal * cm * ΔTmetal
Substituting the given values:
84 g * 4.184 J/g*⁰C * (52.1°C - 20°C) = 68.6 g * cm * (52.1°C - 100°C)
Now, we can solve for cm, which is the specific heat of the metal.
(cm * 68.6 g * (52.1°C - 100°C)) = (84 g * 4.184 J/g*⁰C * (52.1°C - 20°C))
cm = (84 g * 4.184 J/g*⁰C * (52.1°C - 20°C))/(68.6 g * (52.1°C - 100°C))
Evaluating the expression:
cm = (3584.784 J)/(68.6 g * (-47.9°C))
cm ≈ -0.998 J/g*⁰C
Since specific heat cannot be negative, we must take the absolute value:
cm ≈ 0.998 J/g*⁰C
Therefore, the specific heat of the metal is approximately 0.998 J/g*⁰C, which can be rounded to 1.0 J/g*⁰C.
Among the given options, the correct answer is (c) 1.717.
mass metal x specific heat metal x (Tfinal-Tinitial) + mass H2O x specific heat H2O x (Tfinal-Tinitial) = 0
You know all of these except specific heat of the metal. Substitute and solve for that. Post your work if you get stuck.