An object with a mass of $120 g$ is dropped into $640 mL$ of water at $0^@C$ . If the object cools by $48 ^@C$ 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 · 2018-01-29T19:55:36

0.33 cal/gºC

At equilibrium, the final temperature $T$ of both the object and the mass of water must be the same. In addition, in this condition:

$sumQ = 0$ ,

where $Q$ indicates the amount of heat that is absorbed/lost by the two bodies in the system.

Since $Q = mcDeltaT$ , then:

$Q_(object) + Q_(water) = 0$ ;

$m_(object)c_(object)DeltaT_(object) + m_(water)c_(water)DeltaT_(water) = 0$ .

Then:

$120 g * c_(object) * (-48^oC) + 640 g * 1 (cal)/(g^oC) * 3^oC = 0$ ,

where I have used the density of water (1 g/ml) to estimate its mass, and $c_(water) = 1 (cal)/(g^oC).$

Finding the value for $c_(object):$

$c_(object) = (1920)/(5760)(cal)/(g^oC)$ ;

$c_(object) = 0.33 (cal)/(g^oC)$ , approximately.

An object with a mass of 120 g120 g is dropped into 640 mL640 mL of water at 0^@C0^@C. If the object cools by 48 ^@C48 ^@C and the water warms by 3 ^@C3 ^@C, what is the specific heat of the material that the object is made of?