An object with a mass of #120 g# is dropped into #500 mL# of water at #0^@C#. If the object cools by #30 ^@C# and the water warms by #16 ^@C#, what is the specific heat of the material that the object is made of?

Answer 1

The specific heat is #=9.30kJkg^-1K^-1#

The heat is transferred from the hot object to the cold water.

For the cold water, # Delta T_w=16ºC#
For the object #DeltaT_o=30ºC#
# m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)#
#C_w=4.186kJkg^-1K^-1#
Let, #C_o# be the specific heat of the object
#0.120*C_o*30=0.5*4.186*16#
#C_o=(0.5*4.186*16)/(0.120*30)#
#=9.30kJkg^-1K^-1#
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Answer 2

Use the formula ( Q = mc\Delta T ) where ( Q ) is the heat transferred, ( m ) is the mass, ( c ) is the specific heat, and ( \Delta T ) is the change in temperature. Rearrange the formula to solve for ( c ). The specific heat of the material is ( 0.63 , \text{J/g}^\circ\text{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.

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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