A 23.9 g piece of metal heated to 97.8 oC is placed in 52.4 g water at 21.9 oC. After the
metal is added, the temperature of the water rises to 29.9 oC. Calculate the specific heat of the
metal. Express your answer in the units of cal/goC.
3783.63
37342
To calculate the specific heat of the metal, we can use the formula:
q = m × c × ΔT
Where:
q is the heat exchanged
m is the mass of the substance
c is the specific heat of the substance
ΔT is the change in temperature
In this case, the heat gained by the water can be calculated as:
q_water = m_water × c_water × ΔT_water
The heat lost by the metal can be calculated as:
q_metal = m_metal × c_metal × ΔT_metal
Since the heat gained by the water is equal to the heat lost by the metal, we can set up the equation:
q_water = q_metal
m_water × c_water × ΔT_water = m_metal × c_metal × ΔT_metal
We are given:
m_water = 52.4 g
c_water = 1 cal/g°C
ΔT_water = 29.9°C - 21.9°C = 8°C
m_metal = 23.9 g
ΔT_metal = 29.9°C - 97.8°C = -67.9°C (negative because the metal loses heat)
By plugging in the values, we get:
52.4 g × 1 cal/g°C × 8°C = 23.9 g × c_metal × -67.9°C
Now, let's solve for c_metal:
52.4 cal/°C = 23.9 × -67.9 × c_metal
52.4 cal/°C = -1624.1 × c_metal
c_metal = 52.4 cal/°C / -1624.1
c_metal = -0.0322 cal/g°C
Therefore, the specific heat of the metal is approximately -0.0322 cal/goC.
heat lost by hot 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
Substitute the numbers and solve for the only unknown in the equation.