Asample of water with a mass of 587.00 kg is heated with 87 kj of energy to a temperature of 518.4 k. the specific heat of water is 1 j-1 kg k-1. what is the initial temperature of the water?

Answer : Initial temperature of the water is 518.3 K

Explanation :

The specific heat of water (Cp) can be expressed in 2 ways

1) Cp = 1 J g⁻¹ K⁻¹ or

2) Cp = 1 kJ kg⁻¹ K⁻¹

The amount of heat absorbed by water is given by the following equation.

Q = m×Cp×ΔT

Here Q is the amount of energy absorbed in kilojoules

m is the mass of water in kilograms

Cp is specific heat of water

ΔT is the difference is temperature of water.

We have been given that

Q = 87 kJ

Cp = 1 kJ kg⁻¹ K⁻¹

m = 587.00 kg

Let us plug in these values in equation for Q

Q = mxCpxΔT

87 kJ = 587.00 kg x 1 kJ kg⁻¹ K⁻¹ x ΔT

87/587.00 = ΔT

ΔT = 0.15 K

ΔT is calculated as Tf - Ti where Tf is the final temperature and Ti is the initial temperature .

We know that Tf = 518.4 K

And we calculated ΔT as 0.15 K.

Let us plug these values in the equation

ΔT = Tf - Ti

0.15 K = 518.4 K - Ti

Ti = 518.25 K

Initial temperature of the water is 518.3 K

why do the isotopes of a particular element behave differently in nuclear reactions but the same in chemical reactions? chemical reactions involve changes to electrons in an atom; isotopes do not differ in electrons, and so they react the same way during chemical changes.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"