Question

A silver block, initially at 57.2 ∘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 26.5 ∘C. Part A What is the mass of the silver block? Express your answer to two significant figures and include the appropriate units. mm =

Answer 1

Answer:

96.5 grams is the mass of the silver block.

Explanation:

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

$-Q_1=Q_2$

Mass of iron = $m_1=?$

Specific heat capacity of silver= $c_1=0.240 J/g^oC$

Initial temperature of the silver= $T_1=120^oC$

Final temperature = $T_2=T=26.5^oC$

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

Mass of water= $m_2=100.0 g$

Specific heat capacity of water= $c_2=4.184 J/g^oC$

Initial temperature of the water = $T_3=24.8 ^oC$

Final temperature of water = $T_2=T=26.5^oC$

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

$-Q_1=Q_2$

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

On substituting all values:

$-(m_1\times 0.240 J/g^oC\times (26.5^oC-57.2^oC))=100.0 g\times 4.184 J/g^oC\times (26.5^oC-24.8^oC)$

we get, $m_1 = 96.5 g$

96.5 grams is the mass of the silver block.