A piece of metal (75 g at 98.0°C) was dropped in water (95 g at 22.0°C) and the final temperature was measured to be 31.2°C. If the specific heat of water is 4.18 J/g°C. What is the heat capacity of the metal?
heat lost by metal + heat gained by water = 0
[mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial) = 0
Solve for specific heat metal
Note that you have calculated the specific heat capacity of the metal. The problems asks for heat capacity so you must convert specific heat capacity (that's J/gram*C) to heat capacity (that J/C)
To find the heat capacity of the metal, we can use the formula:
Q = mcΔT
where:
Q is the heat transferred
m is the mass of the metal
c is the specific heat capacity of the metal
ΔT is the change in temperature
First, let's calculate the heat transferred to the water using the same formula:
Q_water = m_water * c_water * ΔT_water
where:
Q_water is the heat transferred to the water
m_water is the mass of the water
c_water is the specific heat capacity of water
ΔT_water is the change in temperature of the water
Given:
m_water = 95 g
c_water = 4.18 J/g°C
Initial temperature of water (T1) = 22.0°C
Final temperature of water (T2) = 31.2°C
Calculating ΔT_water:
ΔT_water = T2 - T1
ΔT_water = 31.2°C - 22.0°C
ΔT_water = 9.2°C
Now we can substitute the values into the formula to find Q_water:
Q_water = (95 g) * (4.18 J/g°C) * (9.2°C)
Q_water = 3,997.72 J (rounded to two decimal places)
Now, let's find the heat transferred to the metal:
Q_metal = m_metal * c_metal * ΔT_metal
where:
Q_metal is the heat transferred to the metal (we want to find this)
m_metal is the mass of the metal (given: 75 g)
c_metal is the specific heat capacity of the metal (we want to find this)
ΔT_metal is the change in temperature of the metal (T_metal_final - T_metal_initial)
Given:
m_metal = 75 g
Initial temperature of the metal (T_metal_initial) = 98.0°C
Final temperature of the metal (T_metal_final) = 31.2°C
Calculating ΔT_metal:
ΔT_metal = T_metal_final - T_metal_initial
ΔT_metal = 31.2°C - 98.0°C
ΔT_metal = -66.8°C
We need to convert the negative value of ΔT_metal to positive because we are only interested in the magnitude of the change in temperature, not the direction.
Q_metal = (75 g) * (c_metal) * (66.8°C) (Note: we use 66.8 instead of -66.8 since we want the magnitude)
Q_metal = 5,010 g°C * c_metal
Now, let's solve for c_metal by equating the heat transferred to the water and the heat transferred to the metal:
Q_water = Q_metal
3,997.72 J = 5,010 g°C * c_metal
Now, divide both sides of the equation by 5,010 g°C:
c_metal = 3,997.72 J / 5,010 g°C
c_metal ≈ 0.798 J/g°C (rounded to three decimal places)
Therefore, the heat capacity of the metal is approximately 0.798 J/g°C.