An object with a mass of $21 g$ is dropped into $250 mL$ of water at $0^@C$ . If the object cools by $160 ^@C$ and the water warms by $12 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2017-07-12T10:07:27

$3733.0374Jkg^-1°C^-1$

$m=21g=0.021kg$
$V_w=250ml=250cm^3=250*(10^-2)^3m^3=2.5*10^-4m^3$
$T_(wi)=0°C$
$T_(wf)=12°C$
$Deltatheta_(m)=160C$

Symbols:
$m=$ mass
$v=$ volume
$T=$ temperature
$Deltatheta=$ change in temperature
$Q=$ Heat
$c=$ specific heat capacity

Subscript:
$w=$ water
$i=$ initial
$f=$ final
$m=$ object (a mass)

Heat lost from the object = Heat gained by the water
assuming no heat lost to surrounding
$DeltaQ_w=DeltaQ_m$

$Q=mcDeltatheta$
$m_wc_wDeltatheta=m_mc_mDeltatheta$

density of water $=1gcm^-3=1000kgm^-3$
mass of water, $m_w=1000*2.5*10^-4=0.25kg$
specific heat capacity of water, $c_w=4181Jkg^-1°C^-1$

$0.25*4181*12=0.021*c_m*160$
$12543=3.36*c_m$
$c_m=12543/3.36~~3733.0374Jkg^-1°C^-1$

An object with a mass of 21 g21 g is dropped into 250 mL250 mL of water at 0^@C0^@C. If the object cools by 160 ^@C160 ^@C and the water warms by 12 ^@C12 ^@C, what is the specific heat of the material that the object is made of?