Question

If 832J of energy is required to raise the temperature of a sample of aluminum from 20.0°C to 97.0°C, what mass is the sample of aluminum? (The specific heat of aluminum is 0.90 J/(g × °C).) 0.10 g 10.0 g 12.0 g 57.7 g

Answer 1

Hello!

If 832J of energy is required to raise the temperature of a sample of aluminum from 20.0°C to 97.0°C, what mass is the sample of aluminum? (The specific heat of aluminum is 0.90 J/(g × °C).

a) 0.10 g

b) 10.0 g

c) 12.0 g

d) 57.7 g

Data:

Q (Amount of heat) = 832 J

m (mass) = ?

c (Specific heat) = 0.90 J/(g × ° C)

T (final) = 97 ºC

To (initial) = 20 ºC

ΔT = T - To → ΔT = 97 - 20 → ΔT = 77 ºC

Formula:

Q = m*c*ΔT

Solving:

`Q = m*c*\Delta{T}`

`832 = m*0.90*77`

`832 = 69.3\:m`

`69.3\:m = 832`

`m = (832)/(69.3)`

`\boxed{\boxed{m \approx 12.00\:g}}\end{array}}\qquad\quad\checkmark`

Answer:

12.0 g

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I Hope this helps, greetings ... Dexteright02! =)

Answer 2

Given:
The energy required is
Q = 832 J
The temperature rise is
ΔT = 97 - 20 C
= 77 C = 77 K
The specific heat is
c = 0.9 J/(g-C)
= (0.9 J/(g-K)*(1000 g/kg)
= 900 J/(kg-K)

Let the mass be m kg.
Then
(m kg)*(c J/(kg-K))*(ΔT K) = (Q J)
or
900*77*m = 832
m = 0.012 kg = 12 g

Answer: The mass of aluminum is 12.0 g