Question

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)

Answer 1

Answer:

C on edge

Explanation:

Answer 2

Answer: 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