An object with a mass of 120 g is dropped into 800 mL of water at 0^@C. If the object cools by 30 ^@C and the water warms by 5 ^@C, what is the specific heat of the material that the object is made of?

Jan 20, 2018

1.11 Calorie $g {m}^{-} 1 {C}^{-} 1$

Explanation:

Given,volume of water $800 m l$,we know density of water is $1 \frac{g m}{c {m}^{3}}$ ,so it's mass will be $800 g m$

Suppose,the specific heat of the object be $s$,then using law of thermal equilibrium we can write,

 120×30×s = 800×5×1

Solving we get, $s = 1.11$ CGS units

Jan 20, 2018

$C {p}_{o} = \frac{4.64 J}{g {\cdot}^{o} C}$

Explanation:

The important concept here is that $\text{Heat lost=Heat gained}$

Since the mass of the water is not given, we can derive it through its volume at a given temperature as provided. Knowing these, mass of the water can be calculated using the density formula; i.e.,

$D e n s i t y \left(\rho\right) = \left(\text{mass"(m))/("Volume} \left(V\right)\right)$
$m = \rho \times V$

where:
$V = 800 m l \text{ at } {0}^{o} C$ https://hypertextbook.com/facts/2007/AllenMa.shtml

$m = \frac{0.9998 g}{\cancel{m l}} \times 800 \cancel{m l}$
$m = 799.84 g$

When the object was submerged in the water, temperature of the water has increased to 5^o that signifies heat $\left(Q\right)$ has been absorbed and that can be computed as follows:

$Q = m C p \Delta T$

where:

${m}_{{H}_{20}} = 799.84 g$
${T}_{i} = {0}^{o} C$
${T}_{f} = {5}^{o} C$
$C {p}_{w} = \frac{4.18 J}{g {\cdot}^{o} C}$

$Q = 799.84 \cancel{g} \times \frac{4.18 J}{\cancel{\left(g \cdot o C\right)}} \times 5 \cancel{{\cdot}^{o} C}$
$Q = 16 , 716.66 J$

Remember that $\text{Heat lost" = "Heat gained} = 16 , 716.66 J$. So that, heat capacity of the object $\left(C {p}_{o}\right)$ of this process can be computed with reference to this value of Q. Rearrange formula to isolate the $C {p}_{o}$; that is,

$Q = m C p \Delta T$
$C p = \frac{Q}{m \Delta T}$

where:

$m = 120 g$
$\Delta T = {30}^{o} C$

Cp=(16,716.66J)/(120gxx30^oC
$C p = \frac{4.64 J}{g {\cdot}^{o} C}$

Jan 20, 2018

The specific heat is $= 4.65 k J k {g}^{-} 1 {K}^{-} 1$

Explanation:

The heat is transferred from the hot object to the cold water.

For the cold water,  Delta T_w=5ºC

For the object DeltaT_o=30ºC

${m}_{o} {C}_{o} \left(\Delta {T}_{o}\right) = {m}_{w} {C}_{w} \left(\Delta {T}_{w}\right)$

The specific heat of water ${C}_{w} = 4.186 k J k {g}^{-} 1 {K}^{-} 1$

Let, ${C}_{o}$ be the specific heat of the object

The mass of the object is ${m}_{o} = 0.12 k g$

The mass of the water is ${m}_{w} = 0.800 k g$

$0.12 \cdot {C}_{o} \cdot 30 = 0.800 \cdot 4.186 \cdot 5$

${C}_{o} = \frac{0.800 \cdot 4.186 \cdot 5}{0.12 \cdot 30}$

$= 4.65 k J k {g}^{-} 1 {K}^{-} 1$