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.

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