when 49.5 J of heat was transferred to 7.3 g of iron at 22 degrees celius the temp. of the iron increases to 37 degrees celcius. What is the specific heat capacity of iron in J/g degrees Celcius?
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Convert °C to Kelvin (K)......22°C + 273 = 295K and 37°C + 273 = 310K.
mcΔT = Q.
7.3g x c x 15K = 49.5J
c (Specific Heat Capacity) = 49.5 / (7.3 x 15) = 49.5 / 109.5 = 0.452J/g/°C Specific Heat of Iron.
mcΔT = Q.
7.3g x c x 15K = 49.5J
c (Specific Heat Capacity) = 49.5 / (7.3 x 15) = 49.5 / 109.5 = 0.452J/g/°C Specific Heat of Iron.