0.2 kg of brass at 100°C is dropped into 0.8 kg of water at 20°C, their common temperature comes to be 20°C. Calculate the specific heat capacity of brass?
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m1c1(t1-t) = m2c2(t-t2);
{m1= 0.2kg,
m2= 0.8kg,
c2= 4200J/kgK,
t1=100°C,
t2=20°C and t = 20°C}
0.2×c1×(100-20)=0.8×4200 ( as water remains in the same temp. We needn't mention t-t2)
= c1 = 8×420/8×2
c1 = 210 J/kgK
Hope it helps
{m1= 0.2kg,
m2= 0.8kg,
c2= 4200J/kgK,
t1=100°C,
t2=20°C and t = 20°C}
0.2×c1×(100-20)=0.8×4200 ( as water remains in the same temp. We needn't mention t-t2)
= c1 = 8×420/8×2
c1 = 210 J/kgK
Hope it helps
Dårviñ714:
But the specific heat capacity of brass is 380 J/kgK
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