How much heat must be removed from a 120 g block of copper to lower its temperature from 150°C to 20°C? The sp

How much heat must be removed from a 120 g block of copper to lower its temperature from 150°C to 20°C? The specific heat capacity of copper is 0.093 cal/g·C°.


cal

How much heat must be removed from a 120 g block of copper to lower its temperature from 150°C to 20°C? The sp
Q=m·c·(ΔT)


heat in Joules=Q


mass in grams = m


c = specific heat capacity


ΔT= (Final Temperature - Initial Temperature)





Coppers specific heat capacity is listed as


0.385 J/g·C°


So laying out the equations gives us


Q=120g of Cu x 0.385J/g·C° x (20°C - 150°C)


or (-130°C)


grams and C° cancel during multiplication leaving only Joules...





Q = -6006 J





1 cal (calorie) = 4.1840 J (Joules)





I got this from google's calculator, although some sites list it as just 4.18 or 4.1860 and or even 4.1868. So like anything in chemistry the answer you get depends on the values you choose from the start. Just because a value has more significant values present doesn't mean its more correct though, some rounding has always been involved like it or not...





okay so just using what I got it is then found that


-6006 J x 1 cal/4.1840 J = -1435.47 cal





so about 1435 calories will have to be removed to change the temperature of120g of Copper from 150°C to 20°C.