You have questions? We got answers!

If the specific heat of lead is 0.129 J/g °C, how much heat is given off if the temperature of 135 g of lead drops from 100.0 °C to 36 .0°C?

Science & Mathematics by Anonymous 2018-05-11 16:11:27

Social Science

If the specific heat of lead is 0.129 J/g °C, how much heat is given off if the temperature of 135 g of lead drops from 100.0 °C to 36 .0°C?

5 answers

  • Anonymous

    Q = mc(T2 - T1) Q = 135 x0.129 x (100 - 36) Q = 1114.56J = 1.11kJ

  • Anonymous

    The following equation is used to calculate the amount of heat that is given off. Q = mass * specific heat * (Ti – Tf) Q = 135 * 0.129 * (100 – 36) = 1,114.56 joules You can round this number to three significant digits. This is 1,110 joules.

  • Anonymous

    will this show

  • Anonymous

    watch the units... E = 0.129 J/g°C x 135g x (100–36)C = 1115 J

  • Anonymous

    Remember the equation Q = mcDeltaT Q = heat /J m = mass / 135 g c = 0.129 J/gCo DeltaT = 36 - 100 = - 64 Hence Q = 135 g X 0.129 J/gCo X -64Co Q = -1114.56 J = -1.11456kJ

ReCapcha
Not a bot