# Apiece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °c and dropped

Apiece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °c and dropped into 84 g of water (with an initial temperature of 20 °c) in a calorimeter. the final temperature of the system is 52.1°c. the specific heat of water is 4.184 j/g*⁰c. what is the specific heat of the metal?
a) 0.171
b) 0.343
c) 1.717
d) 3.433

## This Post Has 8 Comments

The specific heat of the metal comes out to be $3.423J/g^oC$

Explanation: Heat absorbed or heat lost, q is written as:

$q=mc\Delta T$

We are given that an unknown metal is dropped in water. So, in this process:

Heat lost by the metal = Heat gained by the water

Mathematically,

$(mc\Delta T)_m=-(mc\Delta T)_w$     ....(1)

For metal:

$T_2=52.1^oC$

$T_1=100^oC$

$\Delta T=T_2-T-1=(52.1-100)^oC=-47.9^oC$

m = 68.6 g

c = ?

For Water:

$T_2=52.1^oC$

$T_1=20^oC$

$\Delta T=T_2-T-1=(52.1-20)^oC=32.1^oC$

m = 84 g

$c=4.184J/g^oC$

Putting this in equation 1, we get

$68.6g\times c_m\times (-47.9^oC)=84g\times 4.184J/g^oC\times 32.1^oC$

$c_m=3.423j/g^oC$

2. Bassoonist says:

The heat absorbed by water will be equal to heat given by the metal

The heat absorbed by water = Q = msΔt

Where

m = mass of water

s = specific heat of water = 4.18 J g⁻¹°C⁻¹

Δt = final temperature - initial temperature = (52.1-20)°C = 32.1°C

So heat absorbed by water = 84 X 4.184 X 32.1 = 11281.74 J

Heat given by metal

= mass of metal X specific heat of metal X change in temperature

heat given by metal = 68.6 X specific heat of metal X (100-52.1)

11281.74 J = 68.6 X specific heat of metal X (100-52.1)

specific heat of metal = 3.43 J g⁻¹°C⁻¹

3. dri16947 says:

Letter b 0.343 tis the respost

4. MrSavannahCat says:

52.1-4.184=47.916 You then either add or subtract this from 68.6 the first temperature of the metal. This will give you your total.

5. karencbetancourt says:

Option C. Hope I helped :)!

6. tefanyc13 says:

Option (d)

Explanation:

Let c be the specific heat of the metal.

By use of principle of caloriemetry,

Heat lost by the hot body = Heat gained by the cold body

Heat lost by metal = Heat gained by water

mass of metal x specific heat of metal x fall in temperature of metal = mass of water x specific heat of water x rise in temperature of water

68.6 x c x (100 - 52.1) = 84 x 4.184 x (52.1 - 20)

3285.94 x c = 11281.7376

c = 3.433 J/g C

7. gungamer720 says:

The piece of unknown metal is in thermal equilibrium with water such that Q of metal is equal to Q of the water. We write this equality as follows:

-Qm = Qw
Mass of metal (Cm)(ΔT) = Mass of water (Cw) (ΔT)

where C is the specific heat capacities of the materials.

We calculate as follows:

-(Mass of metal (Cm)(ΔT)) = Mass of water (Cw) (ΔT)
-68.6 (Cm)(52.1 - 100) = 42 (4.184) (52.1 - 20)
Cm = 1.717 > OPTION C

8. andrew8228 says:

Q= (m)(cp) (delta T)

cp=-q/23.8  (32.5-100.0) ( q is negative because heat was lost)