HELP.
A piece of tin (specific heat 0.222 J/g°C) with an initial temperature of 80.0°C is placed in 100.0g of water (specific heat of 4.184 J/g°C) with an initial temperature of 30.0°C. If the system reaches a final temperature of 34.0°C, what is the mass of the piece of tin?

164g

603g

1.92g

100.g

The mass of tin is 164 grams

Explanation:

Step 1: Data given

Specific heat heat of tin = 0.222 J/g°C

The initial temeprature of tin = 80.0 °C

Mass of water = 100.0 grams

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

Initial temperature = 30.0 °C

The final temperature = 34.0 °C

Step 2: Calculate the mass of tin

Heat lost = heat gained

Qlost = -Qgained

Qtin = -Qwater

Q = m*c*ΔT

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

⇒with m(tin) = the mass of tin = TO BE DETERMINED

⇒with c(tin) = the specific heat of tin = 0.222J/g°C

⇒with ΔT(tin) = the change of temperature of tin = T2 - T1 = 34.0°C - 80.0°C = -46.0°C

⇒with m(water) = the mass of water = 100.0 grams

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

⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 34.0° C - 30.0 °C = 4.0 °C

m(tin) * 0.222 J/g°C * -46.0 °C = -100.0g* 4.184 J/g°C * 4.0 °C

m(tin) =  163.9 grams ≈ 164 grams

The mass of tin is 164 grams