Physics Lead Mass

Question

A 24-g ice cube floats in 220 g of water in a 100-g copper cup; all are at a temperature of 0°C. A piece of lead at 86°C is dropped into the cup, and the final equilibrium temperature is 12°C. What is the mass of the lead? (The heat of fusion and specific heat of water are 3.33  10⁵ J/kg and 4,186 J/kg · °C, respectively. The specific heat of lead and copper are 128 and 387 J/kg ·°C, respectively.)

______ kg

Answer 1

Hi Lindsey,

A 24-g ice cube floats in 220 g of water in a 100-g copper cup; all are at a temperature of 0°C. A piece of lead at 86°C is dropped into the cup, and the final equilibrium temperature is 12°C. What is the mass of the lead? (The heat of fusion and specific heat of water are 3.33  10⁵ J/kg and 4,186 J/kg · °C, respectively. The specific heat of lead and copper are 128 and 387 J/kg ·°C, respectively.)

Given

1. 24g Ice Cube Initial Temperature T_i 0°C Final Temp T_f 12°C Heat of Fusion 3.33  10⁵ J/kg
2. 220g Water T_i 0°C; T_f 12°C Specific Heat 4,186 J/kg · °C
3. 100 g Copper T_i 0°C; T_f 12°C Specific Heat 387 J/kg ·°C
4. _____ g Lead T_i 86°C; T_f 12°C Specific Heat 128 J/kg ·°C

Find: Mass of Lead in kg? __________

To determine the mass of the Lead, we need to look at the heat energy of this system.

We start out with Ice in a copper cup filled with water. The ice, water and copper are at the same temperature to start. Ice is at a temperature of 0°C. So the water and the copper are at 0°C.

Lead is added and the entire system gets to an equilibrium of 12°C. 1) This means that the ice melted, 2) then the water and cup increased in temperature to 12°C. 3) Finally the lead decreased in temperature to 12°C

If we find the heat energy or q of the system, then we will be able to find the mass of the lead.

When a substance goes through a state change, it stays at the same temp and uses the energy to break bonds (in the case of the melting ice)

The formula for that is q = m·ΔH_f Where H_f is the heat of fusion Note the units are J/kg

1) Ice melting Energy q = m·ΔH_f

q_ice=0.024 g H₂O_(s) x 3.33  10⁵ J/k = 7992 J

2) Energy for water (ice melted plus water in cup) to go from 0 to 12degrees

The equation for this is the q=C_p×m×ΔT C_p = the Specific Heat in J/kg ·°C

q_water=C_p×m×ΔT = (4,186 J/kg · °C)(0.024kg + 0.220kg)_H2O (12°C - 0°C) = 12256.608 J

3) Energy for Copper to go from 0 to 12degrees

q_copper=C_p×m×ΔT = (387 J/kg ·°C)(0.100 kg)(12°C - 0°C) = 464.4 J

4) The energy of the lead will be a negative as it was losing heat. Delta T is final temp - Initial Temp

q_lead = C_p×m×ΔT = (128 J/kg ·°C)(m)(12°C - 86°C) = (-9472 J/kg)(m)

q_ice+q_water + q_{copper +} q_lead = 0

7992 J +12256.608 J + 464.4 J + (-9472 J/kg)(m)=0

20713.008 J = (9472 J/kg) m

2.187 kg