An object with a mass of #32 g# is dropped into #480 mL# of water at #0^@C#. If the object cools by #80 ^@C# and the water warms by #6 ^@C#, what is the specific heat of the material that the object is made of?

Answer 1

Lily Adams

The specific heat is #=4.71kJkg^-1K^-1#

The heat that the cold water absorbs and the heat that is transferred from the hot object are equal.

For the cold water, # Delta T_w=6º#

For the object #DeltaT_o=80º#

# m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)#

#C_w=4.186KJkg^-1K^-1#

#m_0 C_o*80 = m_w* 4.186 *6#

#0.032*C_o*80=0.48*4.186*6#

#C_o=(0.48*4.186*6)/(0.032*80)#

#=4.71kJkg^-1K^-1#

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Answer 2

Emily Abbate

Specific heat (c) = 0.5 cal/(g °C)

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Answer from HIX Tutor

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.