Answer: The initial temperature of the iron was 
Explanation:
As we know that,
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
.................(1)
where,
q = heat absorbed or released
= mass of iron = 360 g
= mass of water = 750 g
= final temperature = 
= temperature of iron = ?
= temperature of water = 
= specific heat of iron = 
= specific heat of water= 
Now put all the given values in equation (1), we get
![-360\times 0.450\times (46.7-x)=[750\times 4.184\times (46.7-22.5)]](https://tex.z-dn.net/?f=-360%5Ctimes%200.450%5Ctimes%20%2846.7-x%29%3D%5B750%5Ctimes%204.184%5Ctimes%20%2846.7-22.5%29%5D)
Therefore, the initial temperature of the iron was
Mg is magnesium. NO3 is nitrate. This gives you magnesium nitrate as an answer.
Answer:
Explanation:
mass = 400 grams * [1 kg/1000 grams] = 0.400 kg
c = 387 Joules / (oC * kg)
Δt = 55 - 30 = 25 oC
E = m*c * Δt
E = 0.4 * 387 * 25
E = 3870 Joules
