# An object with a mass of #32 g# is dropped into #250 mL# of water at #0^@C#. If the object cools by #16 ^@C# and the water warms by #3 ^@C#, what is the specific heat of the material that the object is made of?

Now using the principle of calorimetry

Heat gained by water equals heat lost by an object.

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Use the formula: ( Q = mc\Delta T )

( Q_{object} = -Q_{water} )

( c_{object} = \frac{{m_{object} \cdot \Delta T_{object}}}{{m_{water} \cdot \Delta T_{water}}} )

( c_{object} = \frac{{32 , \text{g} \cdot (-16 , ^\circ \text{C})}}{{250 , \text{g} \cdot 3 , ^\circ \text{C}}} )

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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