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    I'm confused about something my Chemistry teacher told me. When using q=mcΔT, when temperature goes up you have to make it negative but this doeant make sense. I've read that when temperature goes up it is an exothermic reaction, and so ΔH should be negative, but having a positive ΔT (if temperature increases) will result in a positive answer, which suggests it is endothermic.
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    (Original post by C_Hope)
    I'm confused about something my Chemistry teacher told me. When using q=mcΔT, when temperature goes up you have to make it negative but this doeant make sense. I've read that when temperature goes up it is an exothermic reaction, and so ΔH should be negative, but having a positive ΔT (if temperature increases) will result in a positive answer, which suggests it is endothermic.
    Exo and endo is referring to the reactants themselves. If the reaction is exothermic the reactants are losing energy so therefore its surroundings gain energy since energy is conserved so the temp goes up. And vice versa for an endothermic reaction.

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    So what would I do in a calculation? Because delta H would still come out positive when it shouldn't be
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    (Original post by C_Hope)
    So what would I do in a calculation? Because delta H would still come out positive when it shouldn't be
    You just have to know that the energy gained by the water = the energy lost by the reactants due to conservation of energy. Therefore if the water has gained x joules of energy the reactants have lost x joules of energy

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    It's best not to worry about the sign of T and the sign of Q, just state as a decrease or increase. See your specific mark schemes, but from what I can see in mine (Edexcel) they only every give marks for the sign in the final answer for H. Best just to think carefully about how to give your final answer.
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    And so should I just adjust my answer accordingly? Say q= 16J, therefore the water gained 16J and reactants lost 16J, and so delta H for the reactants is -16x10^-3/moles to give me delta H in kJmol^-1
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    (Original post by C_Hope)
    And so should I just adjust my answer accordingly? Say q= 16J, therefore the water gained 16J and reactants lost 16J, and so delta H for the reactants is -16x10^-3/moles to give me delta H in kJmol^-1
    Yep

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    And it would be the same if I were measuring the temperature change of a reaction directly? Say I stick a thermometer in a neutralisation reaction?
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    I teach:

    Delta T = final T - initial T (if T inc, then Delta T = +ve)

    Q = m c Delta T (if T inc, then Q = +ve)

    Delta H = - Q / n (if T inc, then Delta H = -ve)

    The minus sign no longer magically appears.

    For the record, if T dec, then Delta T = -ve, then Q = -ve (mark schemes ignore) and Delta H will be +ve = w00t.
 
 
 
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