Question

A 40.5 g block of an unknown metal is heated in a hot water bath to 100.0°C. When the block is placed in an insulated vessel containing 130.0 g of water at 25.0°C, the final temperature is 28.0°C. Determine the specific heat of the unknown metal. The cs for water is 4.18 J/g°C.

Answer 1

C = 0.5621J/g°C

Step-by-step explanation:

Heat lost by metal = Heat gained by water

Heat gained by water;

H = MC ΔT

ΔT = 28 -25 = 3.0°C

M = 130.0g

C = 4.18 J/g°C

H = 130 * 4.18 * 3 = 1639J

Heat lost my metal;

H = MCΔT

ΔT = 100 - 28 = 72°C

M = 40.5g

C = ?

C = H / MΔT

C = 1639 / (40.5 * 72)

C = 0.5621J/g°C