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 need to use the formula:
Q = m × c × ΔT
Where:
- Q is the heat gained or lost by the substance (in this case, the metal)
- m is the mass of the substance (metal) in grams
- c is the specific heat of the substance in calories per gram per degree Celsius
- ΔT is the change in temperature of the substance (final temperature - initial temperature)
First, let's calculate the heat gained by the water. We will assume that the specific heat of water is 1 calorie per gram per degree Celsius.
Q_water = m_water × c_water × ΔT_water
Given that the mass of water (m_water) is 100 mL and the temperature change (ΔT_water) is 24 - 21 = 3 degrees Celsius, we have:
Q_water = 100 g × 1 cal/g°C × 3°C = 300 calories
Now let's calculate the heat lost by the metal.
Q_metal = m_metal × c_metal × ΔT_metal
Given that the mass of the metal (m_metal) is 30 grams and the temperature change (ΔT_metal) is 100 - 24 = 76 degrees Celsius, and we need to find the specific heat of the metal (c_metal).
We can rearrange the formula to solve for c_metal:
c_metal = Q_metal / (m_metal × ΔT_metal)
Substituting the values, we have:
c_metal = Q_metal / (30 g × 76°C)
Now, we know that the heat gained by the water (Q_water) is equal to the heat lost by the metal (Q_metal). Therefore, Q_water = Q_metal.
Substituting the values:
300 calories = Q_metal
Now, we can solve for c_metal:
c_metal = 300 calories / (30 g × 76°C)
Simplifying the expression gives us:
c_metal = 0.1315 cal/g°C
So the specific heat of the metal is approximately 0.1315 calorie per gram per degree Celsius.
heat lost by metal + heat gained by H2O = 0
[mass metal x specific heat metal x (Tfinal-Tinitial)] +[mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for specific heat metal.