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105 mL 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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Answer : The mass of the steel bar is, 35.2 grams.

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.


q_1=-q_2


m_1* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)

where,


c_1 = specific heat of steel =
0.452J/g^oC


c_2 = specific heat of water =
4.18J/g^oC


m_1 = mass of steel rod = ?


m_2 = mass of water =
Density* Volume=1.00g/mL* 105mL=150g


T_f = final temperature of mixture =
21.3^oC


T_1 = initial temperature of steel =
2.0^oC


T_2 = initial temperature of water =
22.0^oC

Now put all the given values in the above formula, we get


m_1* (0.452J/g^oC)* (21.3-2.0)^oC=-(105g)* 4.18J/g^oC* (21.3-22.0)^oC


m_1=35.2g

Therefore, the mass of the steel bar is, 35.2 grams.

User Bartu
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