When 72 g of a metal a t 97.0°C is added to 100 g of water at 25.0°C the final temperature is
found to be 29.1°C. What is the heat capacity per gram of the metal?
(Heat capacity of H2O = 4.184 J/g - °C )
To find the heat capacity per gram of the metal, we can use the formula:
Q = mcΔT
Where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's calculate the heat transferred for the water using the given values:
Q_water = mcΔT
Q_water = (100 g) * (4.184 J/g-°C) * (29.1°C - 25.0°C)
Q_water = 100 * 4.184 * 4.1
Now, let's calculate the heat transferred for the metal:
Q_metal = mcΔT
Q_metal = (72 g) * (c) * (29.1°C - 97.0°C)
Since the metal and water reach the same final temperature, the heat transferred by the metal is equal to the heat transferred by the water.
Q_metal = Q_water
Therefore,
(72 g) * (c) * (29.1°C - 97.0°C) = 100 * 4.184 * 4.1
Now, we can solve this equation to find the value of c, the heat capacity per gram of the metal.