BLACK’S PRINCIPLE At the end of the chapter, you should be able :

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Presentation transcript:

BLACK’S PRINCIPLE At the end of the chapter, you should be able : To apply formula Q=m.c.∆t; Q=m.U;Q=m.L for solving simple problem.

Black’s Principle The amount of heat energy given by a warm substance is the same as the amount of heat energy received by cooler substance

Suppose there are two substances having different temperatures Mass = m1 Specific heat = c1 Temperature = t1 Substance 2 Mass = m2 Specific heat = c2 Temperature = t2 If t1>t2 , Then substance 1 will give an amount of heat energy to substance 2, so

Heat energy given same with of heat energy received So; Q1 = Q2 m1.c1.∆t1 = m2.c2.∆t2 If we find m1 :

Example 2 kg of water have 20oC is immersed in the container. Which is have 1 kg with 60oC. What is the final temperature ? Solution Given : m1= 2 kg t1=20oC m2= 1 kg t2=60oC Unknown : t(final) ?

2 kg.{t(final)-20oc} = 1 kg.{60oc-(tfinal)} Solution : Q1 = Q2 m1.c1.∆t1 = m2.c2.∆t2 2 kg.{t(final)-20oc} = 1 kg.{60oc-(tfinal)} 2.t(final)-40oc= 60oc- t(final) 3. t(final)= 100 oC t(final)=33.3 oC

Exercise 100 g of iron is heated up to 80oC. The iron is immersed in water at 30oC;then the temperature of the water increases to 32oC. What is the mass of water ? c iron = 450j/kgoC, c water = 4200 j/kgoC 2. Kedalam bak mandi dicampurkan 150 kg air panas bersuhu 90 oC dengan air dingin bersuhu 10 oC sehingga menjadi air hangat bersuhu 40 oC. Hitung massa air dingin yang di campurkan ?