The specific heat of iron is 0.107 cal/g x oC. How much heat is transferred when a 24.7 kg iron ingot is cooled from 880 oC to 13 oC?

heat = 229139.43 cal

Explanation:

given data

specific heat of iron is 0.107 cal/g-°C C = 0.4494 J/g°C;

Mass of iron = 24.7 kg = 24700 g

temperature T1 = 880°C

temperature T2 = 13°C

solution

we know that Heat lost that is express as

heat loss = mcΔT ... 1

here m is the mass and c is specific heat capacity of iron and ΔT is the change in temperature

so here change in the temperature is = 880°C - 13°C

change in the temperature = 867°C

so put here value and we get

heat = 24700 * 0.107 cal/g-°C * 867°C

heat = 229139.43 cal