A 1320 g block of lead initially at 35 degrees Celsius absorbs 3044 j of heat. What is the final temperature of the lead? ( The specific heat of lead is 0.16 j/g degrees Celsius.
Added by Jennifer N.
Step 1
The formula is: q = mcΔT where: q = heat absorbed or released (in joules) m = mass (in grams) c = specific heat capacity (in joules per gram per degree Celsius) ΔT = change in temperature (in degrees Celsius) We are given: q = 3044 J m = 1320 g c = 0.16 J/g°C Show more…
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