Question

Must show work for questions, if possible. (Please also explain, I don't understand)

1. A gas absorbs 0.0 J of heat and then performs 91.3 J of work. The change in internal energy of gas is

2. Two metals of equal mass with different heat capacities are subjected to the same amount of heat. Which undergoes the smallest change in temperature?

3. 30.0 mL of pure water at 280. K is mixed with 50.0 mL of pure water at 340 K. What is the final temperature?​

Answer 1

Answers:

1. D) 91.3 J

2. A) The metal with the higher specific heat capacity

3. B) 318 K

Step-by-step explanation:

1. E, q, w

ΔE = q + w

By convention, anything leaving the system is negative and anything entering the system is positive.

q = 0.0 J

w = -91.3 J

E = 0.0 - 91.3 = -91.3 J

2. Specific heat capacity

q = mCΔT

C = q/(mΔT)

If q and m are the same for each metal, then

ΔT ∝ 1/C = k/C

As C increases, ΔT decreases.

Thus, the metal with the higher specific heat capacity will have the smaller temperature change.

3. Temperature on mixing

There are two heat flows in this problem.

Heat gained by cold water + heat lost by hot water = 0

q₁ + q₂ = 0

m₁CΔT₁ + m₂CΔT₂ = 0

m₁ΔT₁ x + m₂ΔT₂ = 0

Data:

m₁ = 30.0 g

T₁ = 280. K

m₂ = 50.0 g

T₂ = 340 K

Calculations:

ΔT₁ = T_f - Ti = T_f - T₁ = (T_f - 280.) K

ΔT₂ = T_f - Ti = T_f - T₂ = (T_f - 340) K

30.0(T_f - 280.) + 50.0(T_f - 340) = 0

30.0T_f - 8400 + 50.0T_f - 17 000 = 0

80.0T_f - 25 400 = 0

80.0T_f = 25 400

T_f =25 400/80.0

T_f = 318 K

The final temperature of the mixture is 318 K.