Question

A piece of metal weighing 5.10 g at a temperature of 48.6 c was placed in a calorimeter in 20.00 ml of water at 22.1 degrees c the final equilibrium temperature was found to be 29.2 what is teh specific heat of the metal

Answer 1

Heat gained is equal to the heat lost;
Heat lost by metal  = 5.10 ×y ×19.4, where y is the specific heat capacity of the metal.
                              = 98.94 y
Heat gained by water = 20 g× 4.18 j/g× 7.1
                                  = 593.56 Joules
Therefore; 98.94 y = 593.56
                            y = 593.56/98.94
                               = 5.99919 j/g
                               ≈ 6.0 j/g
Therefore, the specific heat capacity of the metal is 6.0 J/g°C

Answer 2

At the final equilibrium temperature, the specific heat capacity of the metal is 5.999 J/g°C.

Given the following data:

• Mass of metal = 5.10 grams

• Final temperature of metal = 48.6°C

• Initial temperature of water = 22.1°C

• Final temperature of water = 29.2°C

• Mass of water = 20.00 ml = 20 grams

• Specific heat capacity of water = 4.18 J/g°C

To find the specific heat capacity of the metal:

Mathematically, quantity of heat is given by the formula;

$Q = mc\theta$

Where:

• Q represents the quantity of heat.

• m represents the mass of an object.

• c represents the specific heat capacity.

• ∅ represents the change in temperature.

The quantity of heat lost by the water = The quantity of heat gained by the metal.

$Q_{lost} = Q_{gained}\\\\mc\theta = mc\theta\\\\20(4.18)(29.2 - 22.1) = 5.10c(48.6 - 29.2)\\\\83.6(7.1) = 5.10c(19.4)\\\\593.56 = 98.94c\\\\c = \frac{593.56}{98.94}$

Specific heat capacity of metal, c = 5.999 J/g°C

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