44.90 g of reactants are placed in a beaker containing 45.00 g of water at 15.00 oC. At the end of the chemical reaction the temperature of the water rises to 25.00 oC. Calculate the energy change of the reaction rounded to the correct number of sig figs.

A

1900 J

B

-1900 J

C

1882 J

D

-1882

E

1884 J

F

-1884 J

PLS help lol

D. -1882J

Explanation:

We can solve the energy released in a chemical reaction in an aqueous medium using the equation:

Q = -m*C*ΔT

Where Q is energy (In J),

m is mass of water (45.00g)

C is specific heat of water (4.184J/g°C)

And ΔT is change in temperature (25.00°C - 15.00°C = 10.00°C)

Replacing:

Q = -45.00*4.184J/g°C*10.00°C

Q = -1882J

Right answer is:

answer: semi-infinite solid initially at a uniform temperature of 400 ◦c is cooled at its surface by convection. the cooling fluid at a constant temperature of 200 ◦c has a convective coefficient (ha) of 250 w/m2.k. the physical properties of the solid: k = 20 w/m.k and α = 4 * 10-5  m2/s. using δx = 0.04 m and δt = 10 s. write the finite difference equations describing the transient heat transfer of the first five δx and the temperature profile after 30 s.