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prohojiy [21]
2 years ago
6

What is the specific heat capacity of an unknown metal if 55.00 g of the metal initially at 90°C is placed in 75mL of water with

an initial temperature of 25°C and the final temperature of the system is 35°C?
Chemistry
1 answer:
Lisa [10]2 years ago
8 0

Answer:

The specific heat capacity of the unknown metal is <u>1.04 J/g°C.</u>

Explanation:

Let the specific heat capacity of the unknown metal be x.

Given:

Mass of the metal is, m=55\ g

Initial temperature of the metal is, T_i=90\ \°C

Volume of water is, V=75\ ml

Specific heat capacity of water is, c_w=4.186\ J/g\°C

Initial temperature of water is, T_{wi}=25\ \°C

Final temperature of the system is, T=35\ \°C

We know that density of water is equal to 1 g/ml.

Mass is given as the product of density and volume.

Therefore, mass of water is given as:

m_w=1\times 75=75\ g

Now, fall of temperature of the unknown metal is given as:

\Delta T_m=T_i-T=90-35=55\ \°C

Rise of temperature of water is given as:

\Delta T_w=T-T_{wi}=35-25=10\ \°C

Now, as per conservation of energy,

Heat lost by metal = Heat gained by water

⇒ mx\Delta T_m=m_wc_w\Delta T_w

Plug in all the given values and solve for x. This gives,

55\times x\times 55=75\times 4.186\times 10\\3025x=3139.5\\x=\frac{3139.5}{3025}=1.04\ J/g\°C

Therefore, the specific heat capacity of the unknown metal is 1.04 J/g°C.

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