An object with a mass of $20 g$ is dropped into $6400 mL$ of water at $0^@C$ . If the object cools by $60 ^@C$ and the water warms by $3 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2018-05-13T06:12:32

Specific heat of the material that the object is made of is $16$

Three things are important here.

We assume no heat is lost in the process. In other words heat lost by the object is equal to heat gained by water.
Heat lost or gained is product of mass of the object / water, fall or gain in its temperature and its specific heat.
Specific heat of water is $1$ . Further mass of water is $1gm.$ per $mL$ i.e. $6400mL$ means $6400$ grams.

Now, if specific heat of the material of object is $S$ , heat lost by the object is $20xx60xxS=1200S$ calories

Heat gained by water is $6400xx3xx1=19200$ calories

Hence $1200S=19200$ i.e. $S=19200/1200=16$

Hence, specific heat of the material that the object is made of is $16$

An object with a mass of 20 g20 g is dropped into 6400 mL6400 mL of water at 0^@C0^@C. If the object cools by 60 ^@C60 ^@C and the water warms by 3 ^@C3 ^@C, what is the specific heat of the material that the object is made of?