- #1

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Thanks

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- Thread starter courtrigrad
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- #1

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Thanks

- #2

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[tex] \Delta E = m C \Delta T [/tex]

what is mass1 and mass2 in your question? and what exactly you wanna find?

- #3

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you are given 50 grams of a metal. When it is placed in a solution the the temperature goes down from 120 degrees to 118 degrees. Hence it is a endothermic reaction. Assuming the specific heat capacity of water is 4.41 kj/mol and we look up the specific heat capacity of the metal, find [tex] \Delta H [/tex] for the entire solution.

would i just use

[tex] S \* c \* \Delta T [/tex] and substitute in the values. Was not sure about this problem. s is specific heat, c is mass, and delta T is change in temperature.

Thanks

would i just use

[tex] S \* c \* \Delta T [/tex] and substitute in the values. Was not sure about this problem. s is specific heat, c is mass, and delta T is change in temperature.

Thanks

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- #4

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[tex]Q=mc\Delta T[/tex]

where Q is enthalpy change, c is specific heat capacity.

Always remember that (heat gained) = -(heat lost) when doing these questions. Decide what has lost and gained heat, then sub in the formula for enthalpy change for each, keeping the correct signs in front of whatever necessary.

- #5

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We know the final temperature of the solution which is 120 degrees. But this is [tex] \Delta T [/tex] only for [tex] H_2O [/tex]. So i solve for [tex] \Delta T [/tex] for the metal and received a value. So do i just subtract those two values? ( [tex] \Delta T_1 - \Delta T_2 [/tex]?

Thanks

Thanks

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- #6

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not sure whether this is [tex] Q [/tex]

Thanks

Thanks

- #7

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If you set up the system you can find the change in temp. of the metal, assuming you have the mass of the water. But, if [itex]\Delta H[/itex] refers simply to enthalpy change, then, as I said ealier, (heat gained) = -(heat lost), meaning you can simply find the Q of the water (negative, since its temp. dropped), and that would be equal in magnitude to the Q for the metal, and therefore for the system. Am I understanding the question correctly?

- #8

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the mass of the water is know. Is it correct if I just solve for [tex] \Delta T_m [/tex] to obtain [tex] \Delta_H [/tex] ?

Because I know [tex] \Delta_T_w [/tex]. So I just set [tex]Q=mc\Delta T[/tex] equal to each other and solve for the other [tex] \Delta_T_m [/tex]

Also I know that [tex] \DeltaH = \Delta H_final - \DeltaH_initial [/tex]

Thanks alot

Because I know [tex] \Delta_T_w [/tex]. So I just set [tex]Q=mc\Delta T[/tex] equal to each other and solve for the other [tex] \Delta_T_m [/tex]

Also I know that [tex] \DeltaH = \Delta H_final - \DeltaH_initial [/tex]

Thanks alot

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- #9

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Are you sure? Temperature is in degrees Celsius, while [itex]\Delta H[/itex] should be in units of energy such as joules or kilojoules. Review my last post. [itex]Q=\Delta H[/itex].

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