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Lera25 [3.4K]
3 years ago
5

A 825 g iron block is heated to 352 degrees C and is placed in an insulated container (of negligible heat capacity) containing 4

0.0 g of water at 20.0 degrees C. What is the final equilibrium temperature of the water and iron? specific heat of water = 4186 J/(kg K), specific heat of water vapor = 2090 J/(kg K), specific heat of iron= 560 J/(kg K), latent heat of vaporization for water= 2.26*10^6 J/kg
Chemistry
1 answer:
Stella [2.4K]3 years ago
8 0

Answer : The final equilibrium temperature of the water and iron is, 537.12 K

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\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron =  560 J/(kg.K)

c_1 = specific heat of water = 4186 J/(kg.K)

m_1 = mass of iron = 825 g

m_2 = mass of water = 40 g

T_f = final temperature of water and iron = ?

T_1 = initial temperature of iron = 352^oC=273+352=625K

T_2 = initial temperature of water = 20^oC=273+20=293K

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

(825\times 10^{-3}kg)\times 560J/(kg.K)\times (T_f-625K)=-(40\times 10^{-3}kg)\times 4186J/(kg.K)\times (T_f-293K)

T_f=537.12K

Therefore, the final equilibrium temperature of the water and iron is, 537.12 K

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