An object with a mass of $64 g$ is dropped into $210 mL$ of water at $0^@C$ . If the object cools by $96 ^@C$ and the water warms by $7 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-07-14T17:53:55

$1.00*10^(-3)kJkg^(-1)K^(-1)$

Object:
Mass - $64g=0.064kg$
Specific heat cpacity - ?
Temperature change = $96K$

Water:
Mass - $210mL=0.00021m^3~~0.00021kg$ .
Specific heat capacity - $4.18 kJ K^(-1) kg^(-1)$
Temperature change = $7K$

$m_oC_oDeltaT_o=m_wC_wDeltaT_w$

$(0.064*96)C_o=(4.18*7*0.00021)$

$C_o=(4.18*7*0.00021)/(0.064*96)=0.00100009766~~1.00*10^(-3)kJkg^(-1)K^(-1)$

An object with a mass of 64 g64 g is dropped into 210 mL210 mL of water at 0^@C0^@C. If the object cools by 96 ^@C96 ^@C and the water warms by 7 ^@C7 ^@C, what is the specific heat of the material that the object is made of?