30 grams of a metal at 100 degrees Celsius is placed in a calorimeter containing 100 mL of water at 21 degrees Celsius. After a few minutes, the temperature inside the calorimeter is 24 degrees Celsius. What is the specific heat of the metal?
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To find the specific heat of the metal, we can use the equation:
Q = mcΔT
Where:
Q = heat gained or lost
m = mass of the substance
ΔT = change in temperature
c = specific heat capacity
In this case, the water is the substance whose heat capacity we know. We can calculate the heat gained by the water using the equation:
Q_water = m_water * c_water * ΔT_water
The heat lost by the metal can be calculated using the equation:
Q_metal = m_metal * c_metal * ΔT_metal
Since energy is conserved, the heat gained by the water is equal to the heat lost by the metal:
Q_water = -Q_metal
Rearranging the equation, we get:
m_water * c_water * ΔT_water = -m_metal * c_metal * ΔT_metal
Now let's plug in the given values:
m_water = 100 mL = 100 g (since 1 mL of water is approximately equal to 1 g)
c_water = specific heat capacity of water = 4.18 J/g°C
ΔT_water = 24°C - 21°C = 3°C
m_metal = 30 g
ΔT_metal = 24°C - 100°C = -76°C (negative because the metal loses heat)
Substituting these values into the equation:
100 g * 4.18 J/g°C * 3°C = -30 g * c_metal * -76°C
Simplifying the equation:
1254 J = 2280 g * c_metal
Dividing both sides of the equation by 2280 g:
c_metal = 1254 J / 2280 g ≈ 0.549 J/g°C
Therefore, the specific heat of the metal is approximately 0.549 J/g°C.