Question

A sample of lead is heated to 189.4 degree Celsius.It is then placed in a cup containing 32.8 g water at 24.3 degree Celsius.The final temperature of the water and lead is 29.6 degree Celsius.Calculate the mass of the sample of lead.The specific heat of water is 4.184 J/ g degree Celsius and the specific heat of lead is 0.130J/ g degree Celsius.

Answer 1

The mass of lead is 35.0 grams

Step-by-step explanation:

Step 1: Data given

Temperature of lead = 189.4 °C

Mass of water = 32.8 grams

Temperature of water = 24.3 °C

The final temperature of the water and lead is 29.6 degree Celsius.

The specific heat of water = 4.184 J/g°C

The specific heat of lead = 0.130J/ g°C

Step 2: Calculate the mass of lead

Heat lost = heat gained

Qlead = -Qwater

Q =m*c*ΔT

m(lead)*c(lead)*ΔT(lead) = -m(water)*c(water)*ΔT(water)

⇒m(lead) = the mass of lead = TO BE DETERMINED

⇒c(lead) = the specific heat of lead = 0.130 J/g°C

⇒ΔT(lead) = the change of temperature = T2 - T1 = 29.6 °C - 189.4 °C = -159.8 °C

⇒m(water) = the mass of water = 32.8 grams

⇒c(water) = the specific heat of water = 4.184 J/g°C

⇒ΔT(water) = the change of temperature = T2 - T1 = 29.6 °C -24.3 = 5.3 °C

m(lead) *0.130 * -159.8 = -32.8 * 4.184 * 5.3

-m(lead) * 20.774 = -727.35

m(lead) = 35.0 grams

The mass of lead is 35.0 grams