A 0.2 kg piece of unknown metal is heated to 87 °C and then dropped into a 0.28 kg sample of water at 22°C. If the final temperature of the mixture is 24°C, what is the specific heat of the unknown metal?

Answer 1

The specific heat of the metal is #=0.186kJkg^-1K^-1#

The heat that the cold water absorbs from the hot metal is equal to the heat that is transferred from it.

For the cold water, # Delta T_w=24-22=2º#
For the metal #DeltaT_o=87-24=63º#
# m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)#
#C_w=4.186kJkg^-1K^-1#
#m_0 C_o*63 = m_w* 4.186 *2#
#0.2*C_o*63=0.28*4.186*2#
#C_o=(0.28*4.186*2)/(0.2*63)#
#=0.186kJkg^-1K^-1#
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Answer 2

To find the specific heat of the unknown metal, use the heat transfer formula:

[ Q_{\text{metal}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot \Delta T_{\text{metal}} ]

[ Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} ]

[ Q_{\text{lost by metal}} = Q_{\text{gained by water}} ]

[ c_{\text{metal}} = \frac{m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}}{m_{\text{metal}} \cdot \Delta T_{\text{metal}}} ]

Substitute the given values into the formula to find ( c_{\text{metal}} ).

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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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