Question

A 322 g sample of lead (specific heat = 0.138 J/gºC) is placed into 264 g of water at 25°C. If

the system's final temperature is 46°C, what was the initial temperature of the lead?

Answer 1

-6.31°C

Step-by-step explanation:

To solve this problem, we can use the principle of heat transfer:The heat gained by the lead (q_lead) equals the heat lost by the water (q_water).The formula to calculate heat transfer is:

q = m * c * ΔTWhere:

q = heat transfer

m = mass

c = specific heat

ΔT = change in temperatureFor the lead:

q_lead = m_lead * c_lead * ΔT_leadFor the water:

q_water = m_water * c_water * ΔT_waterGiven values:

m_lead = 322 g

c_lead = 0.138 J/gºC

ΔT_lead = T_final - T_initial_lead (unknown)

m_water = 264 g

c_water = 4.18 J/gºC (specific heat of water)

ΔT_water = T_final - T_initial_water = 46°C - 25°C = 21°CSince the heat gained by the lead is equal to the heat lost by the water, we can set up the equation:

m_lead * c_lead * ΔT_lead = m_water * c_water * ΔT_waterSubstituting the given values:

322 g * 0.138 J/gºC * ΔT_lead = 264 g * 4.18 J/gºC * 21°CSimplifying the equation:

44.436 J/ºC * ΔT_lead = 2325.12 J/ºCDividing both sides of the equation by 44.436 J/ºC:

ΔT_lead = 2325.12 J/ºC / 44.436 J/ºC ≈ 52.31°CFinally, we can find the initial temperature of the lead:

T_initial_lead = T_final - ΔT_lead

T_initial_lead = 46°C - 52.31°C ≈ -6.31°CTherefore, the initial temperature of the lead was approximately -6.31°C.