A silver block, initially at 59.4 °C, is submerged into 100.0 g of water at 24.8 °C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 26.2 °C.
What is the mass of the silver block?

Answer :

Answer:

The mass of silver block = 73.95 g

Explanation:

Given,

For water:

Mass = 100.0 g

Initial temperature = 24.8 °C

Final temperature = 26.2 °C

Specific heat of water = 4.184 J/g°C

For silver:

Mass = x g

Initial temperature = 59.4 °C

Final temperature = 26.2 °C

Specific heat of water = 0.2386 J/g°C

Heat gain by water = Heat lost by silver

Thus,  

[tex]m_{water}\times C_{water}\times (T_f-T_i)=-m_{silver}\times C_{silver}\times (T_f-T_i)[/tex]

Where, negative sign signifies heat loss

Or,  

[tex]m_{water}\times C_{water}\times (T_f-T_i)=m_{silver}\times C_{silver}\times (T_i-T_f)[/tex]

So,  [tex]100.0\times 4.184\times (26.2-24.8)=x\times 0.2386\times (59.4-26.2)[/tex]

[tex]x\times \:0.2386\left(59.4-26.2\right)=100\times \:4.184\left(26.2-24.8\right)[/tex]

[tex]x\times \:0.2386\left(59.4-26.2\right)=585.76[/tex]

[tex]x=\frac{585.76}{7.92152}[/tex]

[tex]x=73.95[/tex]

Hence, the mass of silver block = 73.95 g

Lanuel

The mass of the silver block is equal to 73.95 grams.

Given the following data:

  • Initial temperature of silver = 59.4°C
  • Mass of water = 100.0 g
  • Initial temperature of water = 24.8°C
  • Final temperature of water = 26.2°C
  • Final temperature of silver = 26.2°C

Specific heat of silver = 0.2386 J/g°C

Specific heat of water = 4.184 J/g°C

To find the mass of the silver block:

Mathematically, quantity of heat is given by the formula;

[tex]Q = mc\theta[/tex]

Where:

  • Q represents the quantity of heat.
  • m represents the mass of an object.
  • c represents the specific heat capacity.
  • ∅ represents the change in temperature.

The quantity of heat lost by the silver block = The quantity of heat gained by the water.

[tex]Q_{lost} = Q_{gained}\\\\mc\theta = mc\theta\\\\m(0.2386)(59.4 - 26.2) = 100(4.184)(26.2 -24.8)\\\\m(0.2386)(33.2) = 418.4(1.4)\\\\7.9215m = 585.76\\\\m = \frac{585.76}{7.9215}[/tex]

Mass, m = 73.95 grams

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