Answer to Question #87217 in General Chemistry for mike
2019-03-29T02:31:55-04:00
A 14.28-g block of solid tin at 11.10 °C is immersed in a 22.40-g pool of liquid propylene glycol with a temperature of 64.85 °C. When thermal equilibrium is reached, what is the temperature of the tin and propylene glycol?
Specific heat capacities: tin = 0.213 J/g °C; propylene glycol = 2.50 J/g °C
________ °C
1
2019-04-01T07:49:56-0400
Q 1 = − Q 2 Q_1 = -Q_2 Q 1 = − Q 2
Q 1 = c 1 m 1 ( T 2 − T 1 ) = 14.28 g × 0.213 J g × ∘ C × ( T 2 − 11.1 0 ∘ C ) Q_1 = c_1m_1 (T_2 - T_1) = 14.28 g \times 0.213 \frac{J}{g\times ^\circ C}\times (T_2 - 11.10 ^\circ C) Q 1 = c 1 m 1 ( T 2 − T 1 ) = 14.28 g × 0.213 g × ∘ C J × ( T 2 − 11.1 0 ∘ C )
Q 2 = c 2 m 2 ( T 2 − T 1 ) = 22.40 g × 2.50 J g × ∘ C × ( T 2 − 64.8 5 ∘ C ) Q_2 = c_2m_2(T_2 - T_1) = 22.40 g \times 2.50 \frac{J}{g\times ^\circ C} \times (T_2 - 64.85 ^\circ C) Q 2 = c 2 m 2 ( T 2 − T 1 ) = 22.40 g × 2.50 g × ∘ C J × ( T 2 − 64.8 5 ∘ C )
14.28 g × 0.213 J g × ∘ C × ( T 2 − 11.1 0 ∘ C ) = − ( 22.40 g × 2.50 J g × ∘ C × ( T 2 − 64.8 5 ∘ C ) ) 14.28 g \times 0.213 \frac{J}{g\times ^\circ C}\times (T_2 - 11.10 ^\circ C) = - (22.40 g \times 2.50 \frac{J}{g\times ^\circ C} \times (T_2 - 64.85 ^\circ C)) 14.28 g × 0.213 g × ∘ C J × ( T 2 − 11.1 0 ∘ C ) = − ( 22.40 g × 2.50 g × ∘ C J × ( T 2 − 64.8 5 ∘ C ))
T 2 = 62.0 8 ∘ C T_2 = 62.08 ^\circ C T 2 = 62.0 8 ∘ C
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