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4. A 1320 g block of lead initially at 35.0°C absorbs 3044 J of heat. What is the final temperature of the
lead? (The specific heat of lead is 0.16 J/g °C)

Respuesta :

Answer:

49.41° C

Explanation:

DeltaQ = m c T

  3044  = 1320  *  .16   *  T         where T is the temperature change

     T = 14.41

35 + 14.41 =  49.41 C

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