An object with a mass of $30 g$ $30 g$ is dropped into $750 m L$ $750 mL$ of water at $0^{\circ} C$ $0^@C$ . If the object cools by $20^{\circ} C$ $20 ^@C$ and the water warms by $80^{\circ} C$ $80 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2018-04-20T17:56:52

$Delta T_w=80°C$
$Delta T=20°C$
$m=30 g$
$V=750 m L$

$Q_(water)=Q_(object)$
$Q=m c_p Delta T$
$m_w c_p^w Delta T_w=m c_p Delta T$
$c_p=m_w/m (Delta T_w)/(Delta T) c_p^w$

$rho_w=1000 (kg)/m^3$
$c_p^w=4150 J/(kg K)$
$m_w=V* rho_w=750*10^(-3)*10^(-3)m^3*1000 (kg)/m^3=0.75 kg$

$c_p=m_w/m (Delta T_w)/(Delta T) c_p^w=$
$=(0.75 kg)/(0.03 kg) (80+273 K)/(20+273 K) 4150 J/(kg K)=125 (k J)/(kg K)$

An object with a mass of 30 g30g30 g is dropped into 750 mL750mL750 mL of water at 0^@C0∘C0^@C. If the object cools by 20 ^@C20∘C20 ^@C and the water warms by 80 ^@C80∘C80 ^@C, what is the specific heat of the material that the object is made of?