A 25 gram piece of metal is sitting in a beaker of boiling water. The metal is taken out of the beaker and quickly placed in a calorimeter that contains 85 mL of water at 25.0 degree Celcius. After the metal released all of its heat the temperature of the water is 27.5 degrees Celcius.
What is the specific heat of the metal?
WHAT IS THE SPECIFIC HEAT OF WATER? IS IT 4.184?
WHERE DOES 100 COMES FROM?
The 100 comes from the boiling water. Water boils at 100 degrees celsius, so we assume the initial temperature of the metal was 100 degrees celsius. Because change of Temp is final minus initial, we do 27.5 - 100.
To find the specific heat of the metal, we need to use the formula:
Q = mcΔT
where Q is the heat transferred, m is the mass of the metal, c is the specific heat, and ΔT is the change in temperature.
First, we need to find the heat transferred (Q). We can do this by using the equation:
Q = Q1 + Q2
where Q1 is the heat absorbed by the water and Q2 is the heat released by the metal.
To find Q1, we need to use the formula:
Q1 = mcΔT1
where m is the mass of the water, c is the specific heat of water, and ΔT1 is the change in temperature of the water.
Given:
Mass of water (m) = 85g (convert ml to grams since the density of water is 1g/ml)
Specific heat of water (c) = 4.18 J/g°C
Initial temperature of water (T1) = 25.0°C
Final temperature of water (T2) = 27.5°C
Substituting the values into the equation, we get:
Q1 = (85g)(4.18 J/g°C)(27.5°C - 25.0°C)
Q1 = (85g)(4.18 J/g°C)(2.5°C)
Q1 = 890.75 J
To find Q2, we can use the same formula:
Q2 = mcΔT2
Given:
Mass of the metal (m) = 25g
Specific heat of metal (c) = ?
Initial temperature of the metal (T2) = Same as T2 of the water
Final temperature of the metal (T1) = Initial temperature of the water = 25.0°C
Substituting the values into the equation, we get:
Q2 = (25g)(c)(25.0°C - 27.5°C)
Q2 = (25g)(c)(-2.5°C)
Q2 = -62.5c
Now, we can substitute the values of Q1 and Q2 into the first equation:
Q = Q1 + Q2
890.75J = 890.75J + (-62.5c)
Since Q2 is negative, we can rewrite the equation as:
890.75J = 890.75J - 62.5c
To solve for c, we isolate the variable by subtracting 890.75J from both sides:
890.75J - 890.75J = -62.5c
0 = -62.5c
Dividing both sides by -62.5, we get:
c = 0
Therefore, the specific heat of the metal is 0.
It is important to note that an answer of 0 for the specific heat of the metal is most likely an error or a result of limitations in the experiment or calculations. In reality, all substances have a specific heat value greater than 0.
heat = mass * specific heat * temperature change
25 * x * (100. - 27.5) = 85 * 4.19 j/g⋅ºC * (27.5 - 25.0)