Question

A 12.0 g sample of a metal is heated to 90.0 ◦C. It is then dropped into 25.0 g of water. The temperature of the water rises from 22.5 to 25.0 ◦C. The specific heat of water is 4.18 Jg−1◦C −1 . Calculated the specific heat of the metal. Express your answer in Jg−1◦C
I know that the answer is 0.335. My question: Should there be 2 sig figs (0.34) because 25-22.5=2.5? Or is it three sig figs 0.335 because all the original values have 3 sig figs in the beginning.

Answer 1

Answer:

The specific heat of the metal is 0.34 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 12.0 grams

Mass of the water = 25.0 grams

Initial temperature of the metal = 90.0 °C

Initial temperature of the water = 22.5 °C

Final temperature = 25 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of the metal

Qlost = Qgained

Q = m*c*ΔT

Qmetal = -Qwater

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

⇒ with mass of metal = 12.0 grams

⇒ with c(metal) = TO BE DETERMINED

⇒ with ΔT(metal) = T2 - T1  = 25.0°C - 90.0 °C = -65.0 °C

⇒ with mass of water = 25.0 grams

⇒ with c(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.0 - 22.5 °C = 2.5 °C

12.0 * c(metal) * -65.0 °C = -25.0g * 4.184 J/g°C * 2.5°C

-780.0 * c(metal) = -2615  ( 2.6*10^3 with sig figs)

c(metal) = 0.335 (=0.34 with sig figs)

The specific heat of the metal is 0.34 J/g°C