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.
where,
= specific heat of iron = 560 J/(kg.K)
= specific heat of water = 4186 J/(kg.K)
= mass of iron = 825 g
= mass of water = 40 g
= final temperature of water and iron = ?
= initial temperature of iron = 
= initial temperature of water = 
Now put all the given values in the above formula, we get:
Therefore, the final equilibrium temperature of the water and iron is, 537.12 K
Answer:
0.038 g of reactant
Explanation:
Data given:
Heat release for each gram of reactant consumption = 36.2 kJ/g
mass of reactant that release 1360 J of heat = ?
Solution:
As 36.2 kJ of heat release per gram of reactant consumption so first we will convert KJ to J
As we know
1 KJ = 1000 J
So
36.2 kJ = 36.2 x 1000 = 36200 J
So it means that in chemical reaction 36200 J of heat release for each gram of reactant consumed so how much mass of reactant will be consumed if 1360 J heat will release
Apply unity formula
36200 J of heat release ≅ 1 gram of reactant
1360 J of heat release ≅ X gram of reactant
Do cross multiplication
X gram of reactant = 1 g x 1360 J / 36200 J
X gram of reactant = 0.038 g
So 0.038 g of reactant will produce 1360 J of heat.
Answer:
The answer is C) It is given off as heat
Explanation:
