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.

The mass of lead is 35.0 grams

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

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