A sample of an unknown substance has a mass of 0.465 kg. If 3,000.0 J of heat is required to heat the substance from 50.0°C to 100.0°C, what is the specific heat of the substance?
Use q equals m C subscript p Delta T..
0.00775 J/(gi°C)
0.0600 J/(gi°C)
0.129 J/(gi°C)
0.155 J/(gi°C)

0.129J(gioc)

Explanation:

The basic formula for specific heat is Q=MC∆t, where m is the mass, C is the specific heat, and ∆t is the change in temperature.

The parameters given are Q= 3000j = 3kj, mass is 0.465 and ∆t is (100-50)= T2-T1

By imputing them we have;

Q=MC∆T

30= 0.465xCx(100-50)

3=23.25C

C= 3/23.25

=0.129J(gioC

which of the following was specifically permitted by the emancipation proclamation?

free african americans were allowed to fight for the confederacy.

free african americans were allowed to enlist in the union army and navy.

enslaved people in border states were allowed to move to the north.

enslaved people in border states were allowed to become confederate citizens.

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explanation:

it refers to the definition of full-contact sports, where the condition is that the sport must have athletes must "purposely collide with each other