Question

A calorimeter contains 100.0 g of water at 39.8oC. A metal object with a mass of 10.0 g initially at 50.0oC is placed in the water. When equilibrium has been reached, the final temperature of both the water and the metal is 40.0oC. The specific heat capacity of liquid water is 4.184 J/goC. Although there are NO perfectly insulating calorimeters, assume that the container (calorimeter) holding the water does NOT release or absorb heat. This means that all the heat released by the metal equals the heat absorbed by the water.

Which equation below will give (i.e. equals) the Specific Heat of the metal object? Note that only the numerical values (and not units of each factor) are shown.

Answer 1

Answer: First Option: [(100(4.184)(40.0-39.8)] / [(10.0)(50.0-40.0)]

Step-by-step explanation: You can relate each substance to its corresponding data using the formula q = mcΔT. Thus, you can create two separate equations for both the metal object and the water surrounding it.

Water:

q = (100.0)(4.184)(40.0-39.8)

Metal Object:

-q = (10.0)(c)(40.0-50.0)

Since "q" or energy absorbed/released is the same in magnitude for both instances, you can allow "q" to equal one another leaving you with:

(100)(4.184)(40.0-39.8) = (10.0)(c)(50.0-40.0)

Note that the 50.0 and 40.0 "switch" since the energy absorbed by one substance creates an opposite (or negative) reaction for the other, which is why "q" is negative for the metal object as it releases energy.

Isolating "c," the specific heat of the metal object, leaves you with:

[(100(4.184)(40.0-39.8)] / [(10.0)(50.0-40.0)] = c, which is identical to the first option.

Hope this helps!