A sample of nickel is heated to 95.0°C and placed in a coffee-cup calorimeter. If the 50.0 g of water in the calorimeter is initially 22.3°C and warms to 25.4°C, what mass of nickel was heated? (The specific heat of nickel is 0.444 J/°C g and water is 4.18 J/°C g).
A. 15.4 g
B. 57.5 g
C. 2.1 g
D. 21.0 g
To solve this problem, we can use the equation:
q (heat gained by water) = q (heat lost by nickel)
The heat gained by water can be calculated using the formula:
q = m × c × ΔT
where:
q = heat gained (in J)
m = mass of water (in g)
c = specific heat of water (in J/°C g)
ΔT = change in temperature of water (in °C)
Substituting the given values:
m = 50.0 g
c = 4.18 J/°C g
ΔT = (25.4°C - 22.3°C) = 3.1°C
q (heat gained by water) = (50.0 g) × (4.18 J/°C g) × (3.1°C)
q (heat gained by water) = 647.83 J
Now, let's calculate the heat lost by nickel using the same equation:
q (heat lost by nickel) = m × c × ΔT
where:
q = heat lost (in J)
m = mass of nickel (in g)
c = specific heat of nickel (in J/°C g)
ΔT = change in temperature of nickel (in °C)
Substituting the given values:
c = 0.444 J/°C g
ΔT = (25.4°C - 95.0°C) = -69.6°C (negative because heat is lost)
q (heat lost by nickel) = m × 0.444 J/°C g × (-69.6°C)
q (heat lost by nickel) = -30.8256 m J
Since the total heat gained is equal to the total heat lost, we can equate the two equations:
647.83 J = -30.8256 m J
Now we can solve for m:
m = 647.83 J / -30.8256 J/g
m ≈ -21.0 g
Since mass cannot be negative, we need to ignore the negative sign in the calculation and take the absolute value. Therefore, the mass of nickel heated is approximately 21.0 g.
So, the correct answer is option D. 21.0 g.
To solve this problem, we can apply the principle of heat transfer and use the formula:
Q = m * c * ΔT
where:
Q = heat transfer (in joules)
m = mass (in grams)
c = specific heat capacity (in J/°C g)
ΔT = change in temperature (in °C)
We need to calculate the mass of nickel, so let's assign it the variable "m_nickel".
Given:
Initial temperature of water (T1) = 22.3°C
Final temperature of water (T2) = 25.4°C
Mass of water (m_water) = 50.0 g
Specific heat capacity of water (c_water) = 4.18 J/°C g
Specific heat capacity of nickel (c_nickel) = 0.444 J/°C g
The heat transfer for the water can be calculated using the formula:
Q_water = m_water * c_water * ΔT_water
Before we calculate ΔT_water, we need to know the heat transfer for the nickel, which can be calculated using the same formula:
Q_nickel = m_nickel * c_nickel * ΔT_nickel
Since the nickel is heated and then placed in the coffee-cup calorimeter, the heat lost by the nickel is equal to the heat gained by the water, so:
Q_nickel = -Q_water
Since heat lost by the nickel is equal to the heat gained by the water, we can set the equations for heat transfer equal to each other:
m_nickel * c_nickel * ΔT_nickel = -m_water * c_water * ΔT_water
Now, let's calculate ΔT_water and ΔT_nickel:
ΔT_water = T2 - T1 = 25.4°C - 22.3°C = 3.1°C
ΔT_nickel = T_final - T_initial = 25.4°C - 95.0°C = -69.6°C (since the nickel was heated)
Substituting the values into the equation:
m_nickel * 0.444 J/°C g * (-69.6°C) = -50.0 g * 4.18 J/°C g * 3.1°C
Now, we can solve for m_nickel:
m_nickel = (-50.0 g * 4.18 J/°C g * 3.1°C) / (0.444 J/°C g * -69.6°C)
m_nickel = 57.5 g
Therefore, the mass of nickel that was heated is 57.5 g.
So, the correct answer is option B.
heat lost by Ni + heat gained by H2O = 0
[mass Ni x specific heat Ni x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x )Tfinal-Tinitial)] = 0
Substitute and solve for the only unknown, mass Ni.