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creativ13 [48]
3 years ago
15

A sample of an unknown substance has a mass of 89.5 g. If 345.2 J of heat are required to heat the substance from 285 K to 305 K

, what is the specific heat of the substance?
Answer: ____ J/g•K
Physics
2 answers:
Zinaida [17]3 years ago
7 0

Heat gained in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC(T2-T1)
345.2 = 89.5(C)(305 - 285)
C = 0.1928 </span>J/g•K

brilliants [131]3 years ago
6 0

Answer:

0.193 J/g*K on edg

Explanation:

I just took it

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3 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
3 years ago
Help me pls its for my science class i need to show my work
Lostsunrise [7]

Answer:

P = 5880  J

Explanation:

Given that,

The mass of the block, m = 30 kg

The block is sitting at a height of 20 m.

The block will have gravitational potential energy. The formula for gravitational potential energy is given by :

P=mgh\\\\=30\times 9.8\times 20\\\\=5880\ J

So, the required potential energy is equal to 5880  J.

6 0
3 years ago
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