# An object with a mass of 4 g is dropped into 650 mL of water at 0^@C. If the object cools by 12 ^@C and the water warms by 30 ^@C, what is the specific heat of the material that the object is made of?

Mar 9, 2016

C_o = (ΔT_(H_2O))* (M_(H_2O)C_(H_2O))/(M_o*ΔT_o)

${C}_{o} = \frac{{30}^{o} C \cdot 650 g \cdot 4.184 \left(\frac{J g}{c}\right)}{4 g \cdot {12}^{o} C}$

#### Explanation:

Use Heat and the First Law of Thermodynamics. We can use the relationship Q = mcΔT to relate the temperature changes of the object and the water to their masses, specific heats, and the amount of heat supplied to or absorbed by each.
ΔT_o = (ΔQ_o)/(M_oC_o) ===================> (1) Object

ΔT_(H_2O)=(ΔQ_(H_2O))/(M_(H_2O)C_(H_2O)) ===========> (2) ${H}_{2} O$

Divide equation (1) into (2) knowing that (ΔQ_o) = ΔQ_(H_2O)

(ΔT_o)/(ΔT_(H_2O))= (M_(H_2O)C_(H_2O))/(M_oC_o)

Now what the mass of 650 ml of water? The mass of 650 ml of water is $\implies .65 k g$. From a specific heat table
${C}_{{H}_{2} O} = 75.2 \frac{J}{\text{mol}} K$. Now we have all we need, just substitute and solve for ${C}_{o}$

C_o = (ΔT_(H_2O))* (M_(H_2O)C_(H_2O))/(M_o*ΔT_o)

${C}_{o} = \frac{{30}^{o} C \cdot 650 g \cdot 4.184 \left(\frac{J g}{c}\right)}{4 g \cdot {12}^{o} C}$

Now is a question calculating. That said I am a bit suspicious of the answer it is too large for a specific heat value...

Mar 10, 2016

Specific heat of material of object$= 406.25 c a l / g m$

The value is unrealistically high.

#### Explanation:

As the quantities have been given in CGS units, therefore, answer has been worked out in same units.

Value used

Specific heat of water$= 1 c a l / g m = 4.19 k J / k g$

Let the specific heat of the material of object $= {s}_{O}$

Amount of heat exchanged is given as $\Delta Q = m s \Delta t$,
where $m$ is the mass, $s$ is the specific heat and $\Delta t$ is change in temperature.

Heat gained by $650 m L$ of water at ${0}^{o} C$,
assuming $1 m L$ of water $= 1 g m$ of water.

$\Delta {Q}_{g a i n e d} = m s \Delta {t}_{w} = 650 \times 1 \times 30 = 19500 c a l$

Simlarly heat lost by object is given as
$\Delta {Q}_{l o s t} = m {s}_{O} \Delta {t}_{O} = 4 \times {s}_{O} \times 12 = 48 {s}_{O}$

Since, $\Delta H e a {t}_{l o s t} = \Delta H e a {t}_{g a i n e d}$

$\therefore 48 {s}_{O} = 19500$
$\implies {s}_{O} = \frac{19500}{48}$
or ${s}_{O} = 406.25 c a l / g m$