When a 77 g piece of metal with a temperature of 88 Degree Celsius is placed into 277 g of water having a temperature of 20 degree Celsius, the final temperature of the metal and water is 23.3 degree Celsius. What must be the specific heat of the metal?
heat lost by metal + heat gained by water = 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.
Thank you bob. However, what is specific heat of a metal is e metal is unknown.
The problem asks for the specific heat of the metal. That's what you're solving for. Substitute the numbers and solve for specific heat.
[77 x sp.h.metal x (23.3-20.0)] + [277 x 4.184 x (23.3-20.0)] = 0
You have only one unknown; i.e., specific heat metal. Solve for that.
To find the specific heat of the metal, we can use the principle of conservation of energy. The amount of heat lost by the metal will be equal to the amount of heat gained by the water.
The equation for heat transfer is given by:
Q = mcΔθ
Where:
Q is the heat transfer in joules
m is the mass of the substance in grams
c is the specific heat capacity of the substance in J/g°C
Δθ is the change in temperature in °C
In this case, the heat lost by the metal will be equal to the heat gained by the water:
(metal heat lost) = (water heat gained)
The heat lost by the metal can be calculated using the equation:
Qmetal = mcΔθ
The heat gained by the water can be calculated using the equation:
Qwater = mcΔθ
Since the final temperature of the metal and water is the same (23.3°C), we can rewrite the equations as:
Qmetal = mc metal (final temperature - initial temperature)
Qwater = mc water (final temperature - initial temperature)
Since the mass and initial temperature of the water are given, we can calculate its heat gained:
Qwater = (277 g) (4.18 J/g°C) (23.3°C - 20°C)
Now, we can substitute the values into the equation:
Qmetal = Qwater
(mc metal) (23.3°C - 88°C) = (277 g) (4.18 J/g°C) (23.3°C - 20°C)
Simplifying the equation:
(mc metal) (-64.7°C) = (277 g) (4.18 J/g°C) (3.3°C)
Dividing both sides by -64.7°C:
mc metal = [(277 g) (4.18 J/g°C) (3.3°C)] / (-64.7°C)
Calculating the specific heat c metal:
c metal = [(277 g) (4.18 J/g°C) (3.3°C)] / (mc metal)