. 4.) An iron rod at 80°C of mass 0.8 kg is dropped into 1 kg of water at 20°C. What will be the final temperature of rod and water in equilibrium?
Answers
Answer:
Explanation:
Hey,
Lets consider mass of iron as m1=0.8kg
water mass m2=1kg
temperature of iron t2=80*C
temperature of water t2=20*C
According to law of conversation of energy, energy gained = energy lost.
Hence Q1 =Q2
c of iron × m1 × t1-t = c of water × m2 ×t- t2
448 * 0.8 * t1-t = 4186 * 1 * t-t2
Now find answer by yourself I am tired of typing now its just numerical part.
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Answer:
final temperature is 50°C for both water and rod
Explanation:
because when the iron rod is dropped into water,the rod's temperature will decrease and water temperature will increase until they both are at same temperature.the formula is
heat of one object +heat of other object/2
=80+20/2
=100/2=50
note:this is applicable for many questions
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