heat = mass * specific heat * temperature change
25 * x * (100. - 27.5) = 85 * 4.19 j/g⋅ºC * (27.5 - 25.0)
What is the specific heat of the metal?
25 * x * (100. - 27.5) = 85 * 4.19 j/g⋅ºC * (27.5 - 25.0)
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.