105mL of H2O is initially at room temperature (22.0?C). A chilled steel rod at 2

Question

105mL of H2O is initially at room temperature (22.0?C). A chilled steel rod at 2.0?C is placed in the water. If the final temperature of the system is 21.3?C, what is the mass of the steel bar?

Specific heat of water = 4.18 J/g??C

Specific heat of steel = 0.452 J/g??C

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The specific heat of water is 4.18 J/g??C. Calculate the molar heat capacity of water.

molar heat capacity for water =

Explanation / Answer

as density of water = 1g/ml

so,
mass of water = 105 g

so,
using heat gained = heat lost,
m1* s1* delta T1 = m2*s2* delta T2

105* 4.18 * (22-21.3) = m2* 0.452* ( 21.3-2)

solving it ,

mass of steel rod , m2 = 35.2183 ~35.22 g

2.

molar heat capacity = heat required per mole = heat required per 18 g of water

so.
molar heat capacity = 18* specific heat capacity

molar heat capacity = 18* 4.18 = 75.24 J/mol/C .

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