If you have 11.0 g of lead and wish to increase its temperature by 27.0oC, how much heat will -1o -1 have to be added to the lead? ?

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  • 2 weeks ago

    "-1o -1" ?? what does this mean

    skipping that, 

    specific heat lead is 126 J/kgC 

    E = 126 J/kgC x 0.011 kg x 27ºC = 37.4 J

  • 2 weeks ago

    q = mcΔT

    q = 11.0g x 0.129 J/gC x 27.0C = 38.3 J

    38.3 J of heat must be added to 11.0 grams of lead to increase its temperature by 27.0 Celsius degrees.

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