An object with a mass of $50 g$ $50 g$ is dropped into $300 m L$ $300 mL$ of water at $0^{\circ} C$ $0^@C$ . If the object cools by $80^{\circ} C$ $80 ^@C$ and the water warms by $8^{\circ} C$ $8 ^@C$ , what is the specific heat of the material that the object is made of?

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An object with a mass of $50 g$ $50 g$ is dropped into $300 m L$ $300 mL$ of water at $0^{\circ} C$ $0^@C$ . If the object cools by $80^{\circ} C$ $80 ^@C$ and the water warms by $8^{\circ} C$ $8 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-12-07T12:43:20

$C_{o} = 2508 J l k g^{- 1} l K^{- 1}$ $C_o=2508Jcolor(white)(l)kg^(-1)color(white)(l)K^(-1)$

Explanation:

$DeltaE=mcDeltatheta$ , where:

$DeltaE$ = change in energy ( $J$ )
$m$ = mass of the object ( $kg$ )
$c$ = specific heat capacity of the object ( $J$ $kg^(-1)$ $K^(-1)$ )
$Deltatheta$ = change in temperature ( $K$ )

Assuming all heat is transferred we have:
$m_oC_oDeltatheta_o=m_wc_wDeltatheta_w$

$m_o=50g=0.05kg$
$Deltatheta_o=80^circC=80K$
$rho_w~~1000kgcolor(white)(l)m^(-3)$
$V_w=300mL=0.3L=0.0003m^3$
$m_w=1000*0.0003=0.3kg$
$Deltatheta_w=8^circC=8K$
$c_w~~4180Jcolor(white)(l)kg^(-1)color(white)(l)K^(-1)$

$0.05C_o*80=0.3*4180*8$

$C_o=(0.3*4180*8)/(0.05*80)=2508Jcolor(white)(l)kg^(-1)color(white)(l)K^(-1)$

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An object with a mass of $50 g$ $50 g$ is dropped into $300 m L$ $300 mL$ of water at $0^{\circ} C$ $0^@C$ . If the object cools by $80^{\circ} C$ $80 ^@C$ and the water warms by $8^{\circ} C$ $8 ^@C$ , what is the specific heat of the material that the object is made of?

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Explanation:

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1 Answer

Explanation:

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An object with a mass of $50 g$ $50 g$ is dropped into $300 m L$ $300 mL$ of water at $0^{\circ} C$ $0^@C$ . If the object cools by $80^{\circ} C$ $80 ^@C$ and the water warms by $8^{\circ} C$ $8 ^@C$ , what is the specific heat of the material that the object is made of?