Chemistry: A 155.0 g piece of copper at 128 oC is dropped into 250.0 g of water at 17.9 oC. (The specific heat of copper is 0.385 J/goC.) Calculate the final temperature of the mixture. (Assume no heat loss to the surroundings.)

A 155.0 g piece of copper at 128 oC is dropped - id15501285362025, breroyalee2584

Ответ:

Lindsay882

20.01.2022

Chemistry

$T_{eq}=23.85^oC$

Explanation:

Hello,

In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:

$\Delta H_{Cu}=-\Delta H_{H_2O}$

Therefore the equilibrium temperature shows up as:

$m_{Cu}Cp_{Cu}(T_{Cu}-T{eq}) = m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})\\\\T_{eq}=\frac{m_{Cu}Cp_{Cu}T_{Cu}-m_{H_2O}Cp_{H_2O}T_{H_2O}}{m_{Cu}Cp_{Cu}-m_{H_2O}Cp_{H_2O}} \\$

Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:

$T_{eq}=\frac{155.0g*0.385\frac{J}{g^oC}*128^oC+250.0g*4.18\frac{J}{g^oC}*17.9^oC}{155.0g*0.385\frac{J}{g^oC}+250.0g*4.18\frac{J}{g^oC}}=23.85^oC$

Best regards.

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Ответ:

lineaeriksen

10.05.2021

Chemistry

YearPopulation

1 11,500

2 12,075

3 12,679

4 13,313

5 13,978

6 14,677

Explanation:

0,0(0 оценок)

A 155.0 g piece of copper at 128 oC is dropped into 250.0 g of water at 17.9 oC. (The specific heat of copper is 0.385 J/goC.) Calculate the final temperature of the mixture. (Assume no heat loss to the surroundings.)

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