Question

If 20 kg of iron, initially at 12 °C, is added to 30 kg of water, initially at 90 °C, what would be the final temperature of the combined system? (Hint: the heat given up by the water will be equal to the heat gained by the iron) Explain how you would represent this problem in the simulation.

Answer 1

Answer:

final temperature of the combined system T = 84.78°C

Explanation:

Given data

mass of iron ( m1 )   = 20 kg

temperature iron ( t1 ) =  12 °C

mass of water ( m2 ) = 30 kg

temperature of water ( t2 )   =  90 °C

To find out

final temperature of the combined system

solution

we know the energy requirement formula to rise the temp

energy = mass × specific heat  × change in temperature  

we combine both system so both energy will be added

and

we know specific heat of iron ( c1 ) = 0.450 kJ/kg

and specific heat of water ( c2 ) = 4.186 kJ/kg

4.186 joule/gram °C

now combine both energy

energy = mass, m1 × specific heat, c1  × change in temperature, T - t1 + mass, 2 × specific heat, c2  × change in temperature, T - t2

energy = 20 × 0.450  × T - 12  + 30 × 4.186 × T -90

(20)(0.45)(T−12)=(30)(4.186)(90−T)

final temperature of the combined system T = 84.78°C