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?

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Answer 1 · 2017-03-02T06:44:07

The specific heat of the metal is $= 0.186 k J k g^{- 1} K^{- 1}$ $=0.186kJkg^-1K^-1$

The heat transferred from the hot metal, is equal to the heat absorbed by the cold water.

For the cold water, $Delta T_w=24-22=2º$

For the metal $DeltaT_o=87-24=63º$

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

$C_w=4.186kJkg^-1K^-1$

$m_0 C_o*63 = m_w* 4.186 *2$

$0.2*C_o*63=0.28*4.186*2$

$C_o=(0.28*4.186*2)/(0.2*63)$

$=0.186kJkg^-1K^-1$