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? ?
- billrussell42Lv 72 weeks ago
"-1o -1" ?? what does this mean
specific heat lead is 126 J/kgC
E = 126 J/kgC x 0.011 kg x 27ºC = 37.4 J
- pisgahchemistLv 72 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.