A piece of metal with a mass of 32.8 g is heated to 100.5°C and dropped into 138.2 g of water at 20.0°C. The final temperature of the system is 30.2°C. What is the specific heat capacity of the metal?

where: [tex]m_{m}[/tex] is the mass of metal, [tex]c_{m}[/tex] is the specific heat capacity of metal, ΔT is the change in temperature of metal and [tex]m_{w}[/tex] is the mass of water, [tex]c_{w}[/tex] is the specific capacity of water and ΔT is the change in temperature of water.

32.8 x [tex]c_{m}[/tex] x (100.5 - 30.2) = 138.2 x 4.2 x (30.2 - 20.0)

32.8 x [tex]c_{m}[/tex] x 70.3 = 138.2 x 4.2 x 10.2

[tex]c_{m}[/tex] x 2305.84 = 5920.488

[tex]c_{m}[/tex] = [tex]\frac{5920.488}{2305.84}[/tex]

= 2.56761

[tex]c_{m}[/tex] = 2.6 J/g°C

The specific heat capacity of the metal is 2.6 J/g°C.

andromache andromache · Answer 2 · 2022-02-24T13:08:05+01:00

The specific heat capacity of the metal is 2.6 J/g°C.

Calculation of specific heat capacity:

We know that

Heat lost by metal = Heat gained by the water

So,

[tex]m_mc_m\Delta\ T = m_wc_w\Delta\ T[/tex]

Here

[tex]m_m[/tex] is the mass of metal,

[tex]c_m[/tex] is the specific heat capacity of metal,

ΔT is the change in temperature of metal

[tex]m_w[/tex] is the mass of water,

[tex]c_w[/tex] is the specific capacity of water

and ΔT is the change in temperature of water.

Now

32.8 x c_m x (100.5 - 30.2) = 138.2 x 4.2 x (30.2 - 20.0)

32.8 x c_m x 70.3 = 138.2 x 4.2 x 10.2

[tex]c_w[/tex] = 2.6 J/g°C

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