Question

An insulated cup contains 75g of water at 24 degrees. A 26g sample of metal at 82.25 degrees is added. The final temperature of the water and metal is 28.34 degrees. What is the specific heat of the metal?

Answer 1

Heat lost by the metal = Heat gained the  mass of water

Now,
Heat = mass*change in temperature*specific heat capacity

Specific heat of water = 4.18 J/g.°C

Therefore,
75*4.18*(28.34-24) = 26*C*(82.25-28.34)  where C = Specific heat of the metal

C = (75*4.18*4.34)/(26*53.91) = 0.971 J/g.°C

Answer 2

The specific heat of a 26 grams sample of metal at 82.25 degrees is 0.971 J/g°C.

Given the following data:

• Initial temperature of liquid water = 24.0°C

• Final temperature of water = 28.34°C

• Final temperature of metal = 28.34°C

• Mass of water = 75.0 grams

• Mass of metal = 26.0 grams

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

To find the specific heat of the metal:

Mathematically, heat capacity or 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.

Heat lost by water = Heat gained by metal.

$75(4.18)(28.34 - 24) = 26c(82.25 - 28.34)\\\\313.5(4.34) = 26(53.91)c\\\\1360.59 = 1401.66c\\\\c = \frac{1360.59}{1401.66}$

Specific heat of metal, c = 0.971 J/g°C.

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