Question

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 1

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,  

$m_{water}\times C_{water}\times (T_f-T_i)=-m_{silver}\times C_{silver}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,  

$m_{water}\times C_{water}\times (T_f-T_i)=m_{silver}\times C_{silver}\times (T_i-T_f)$

So,  $100.0\times 4.184\times (26.2-24.8)=x\times 0.2386\times (59.4-26.2)$

$x\times \:0.2386\left(59.4-26.2\right)=100\times \:4.184\left(26.2-24.8\right)$

$x\times \:0.2386\left(59.4-26.2\right)=585.76$

$x=\frac{585.76}{7.92152}$

$x=73.95$

Hence, the mass of silver block = 73.95 g