# 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?

Jul 12, 2017

3733.0374Jkg^-1°C^-1

#### Explanation:

$m = 21 g = 0.021 k g$
${V}_{w} = 250 m l = 250 c {m}^{3} = 250 \cdot {\left({10}^{-} 2\right)}^{3} {m}^{3} = 2.5 \cdot {10}^{-} 4 {m}^{3}$
T_(wi)=0°C
T_(wf)=12°C
$\Delta {\theta}_{m} = 160 C$

Symbols:
$m =$mass
$v =$volume
$T =$temperature
$\Delta \theta =$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
$\Delta {Q}_{w} = \Delta {Q}_{m}$

$Q = m c \Delta \theta$
${m}_{w} {c}_{w} \Delta \theta = {m}_{m} {c}_{m} \Delta \theta$

density of water$= 1 g c {m}^{-} 3 = 1000 k g {m}^{-} 3$
mass of water,${m}_{w} = 1000 \cdot 2.5 \cdot {10}^{-} 4 = 0.25 k g$
specific heat capacity of water, c_w=4181Jkg^-1°C^-1

$0.25 \cdot 4181 \cdot 12 = 0.021 \cdot {c}_{m} \cdot 160$
$12543 = 3.36 \cdot {c}_{m}$
c_m=12543/3.36~~3733.0374Jkg^-1°C^-1