A 40g chunk of 90 degree C iron is dropped into a cavity in a very large block of ice at 0 degree C.Find what mass of ice will melt. (The specific heat capacity of iron is 0.11 cal/g degree C)

Answer 1

This problem involves energy transfer by heat. The energy released by the iron as it cools down will go into melting the ice.

First, let's find out how much energy is released as the iron goes from 90°C to 0°C:

q_iron = m_iron * c_iron * ΔT_iron

Where q is the energy, m is the mass, c is the specific heat capacity and ΔT is the difference in temperature.

q_iron = 40g * 0.11 cal/g°C * (0°C - 90°C)
q_iron = 40g * 0.11 cal/g°C * (-90°C)
q_iron = -396 cal

The negative sign signifies that the iron is releasing energy as heat.

Now, let's relate this energy to the amount of ice that will melt. The energy required to melt ice can be expressed as q_ice = m_ice * L_f, where m_ice is the mass of the ice and L_f is the latent heat of fusion for ice (i.e., the energy required to change ice from solid to liquid without any change in temperature). For ice, L_f = 80 cal/g.

Therefore, we can write:

-396 cal = m_ice * 80 cal/g

Solving for m_ice, we obtain:

m_ice = -396 cal / 80 cal/g
m_ice ≈ 4.95 g

So, about 4.95 grams of ice will melt.