Question

A silver block, initially at 58.4∘C, is submerged into 100.0 g of water at 25.0∘C in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 26.7∘C. The specific heat capacities for water and silver are Cs,water=4.18J/(g⋅∘C) and Cs,silver=0.235J/(g⋅∘C).

Answer 1

Answer:

The mass of the silver block is 95.52 grams.

Explanation:

Heat lost by silver will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of silver= $m_1$

Specific heat capacity of silver = $c_1=0.235 J/g^oC$

Initial temperature of the silver = $T_1=58.4^oC$

Final temperature of a silver = $T_2=T=26.7^oC$=

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_1=100.0 g$

Specific heat capacity of water= $c_2=4.186 J/g^oC$

Initial temperature of the water = $T_3=25^oC$

Final temperature of water = $T_3=T=26.7^oC$

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

$-(m_1\times 0.235 J/g^oC\times (26.7^oC-58.4^oC))=100.0 g\times 4.186 J/g^oC\times (26.7^oC-25.0^oC)$

On substituting all values:

we get, $m_1 = 95.52 g$

The mass of the silver block is 95.52 grams.