Question

Calculate the final temperature of 38 mL of ethanol initially at 19 ∘C upon absorption of 552 J of heat. (density of ethanol =0.789 g/mL)

Answer 1

Question #81803, Chemistry, Other

Question

Calculate the final temperature of 38 mL of ethanol initially at 19°C upon absorption of 552 J of heat. (density of ethanol = 0.789 g/mL)

Solution:

According to the conditions:

Volume V(ethanol) = 38 ml

Initial temperature T₀(ethanol) = 19°C

Absorbed heat Q = 552 J

Ethanol density ρ(ethanol) = 0.789 g/mL

Ethanol specific heat c (ethanol) = 2.46 J/g°C

The absorbed heat can be described using the equation:

$$Q = c m \Delta T = c m (T _ {1} - T _ {0})$$

where Q – absorbed heat, c – specific heat, mass, ΔT – change in temperature, T₁ – final temperature, T₀ – initial temperature

Next, ethanol mass can be represented as $m = \rho \times V$.

Therefore:

$$T _ {1} = \frac {Q + c m T _ {0}}{c m} = \frac {Q + c \rho V T _ {0}}{c \rho V} = \frac {5 5 2 J + 2 . 4 6 \mathrm {J / g ^ {\circ} C} \times 0 . 7 8 9 \mathrm {g / m l} \times 3 8 \mathrm {m l} \times 1 9 ^ {\circ} \mathrm {C}}{2 . 4 6 \mathrm {J / g ^ {\circ} C} \times 0 . 7 8 9 \mathrm {g / m l} \times 3 8 \mathrm {m l}} = 2 6. 5 ^ {\circ} \mathrm {C}$$

Answer:

The final temperature is 26.5°C

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