A 150.0-g sample of a metal at 69.7°C is added to 150.0 g of H2O at 10.5°C. The temperature of the water rises to 13.6°C. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water.
[mass metal x specific heat metal x (Tfinal-Tinitial)] + {mass H2O x specific heat H2O x (Tfinal-Tinbitial)] = 0
Substitute all of the numbers and solve for the only unknown; i.e., sp.h. metal.
To calculate the specific heat capacity of the metal, we can use the heat gained by the water and the heat lost by the metal. Here's how you can do it step by step:
1. Calculate the heat gained by the water using the formula:
Qwater = (masswater) * (specific heat capacitywater) * (change in temperaturewater)
In this case:
masswater = 150.0 g
specific heat capacitywater = 4.18 J/g°C (specific heat capacity of water)
change in temperaturewater = (final temperaturewater) - (initial temperaturewater) = 13.6°C - 10.5°C
2. Calculate the heat lost by the metal using the formula:
Qmetal = (massmetal) * (specific heat capacitymetal) * (change in temperaturemetal)
In this case:
massmetal = 150.0 g
specific heat capacitymetal = ?
change in temperaturemetal = (final temperaturemetal) - (initial temperaturemetal) = (final temperaturewater) - (initial temperaturewater) = 13.6°C - 69.7°C
3. Since the heat gained by the water is equal to the heat lost by the metal (assuming no heat is lost to the surroundings), we can set up the equation:
Qwater = Qmetal
4. Substitute the values calculated in steps 1 and 2 into the equation and solve for the specific heat capacity of the metal:
(masswater) * (specific heat capacitywater) * (change in temperaturewater) = (massmetal) * (specific heat capacitymetal) * (change in temperaturemetal)
(150.0 g) * (4.18 J/g°C) * (13.6°C - 10.5°C) = (150.0 g) * (specific heat capacitymetal) * (13.6°C - 69.7°C)
Solve for specific heat capacitymetal.
By plugging in the values and solving the equation, you should be able to find the specific heat capacity of the metal.