Question

J

and at 106°C is placed in 300. g of water at 25.0° C. What will be the
g°C
A 71.0 g piece of metal with specific heat 0.622-
final temperature of the water?
J
9°C
• Use 4.184
for the specific heat of water.

Answer 1

The final temperature of the water will be 9.0°C.

Given that we have a 71.0 g piece of metal with specific heat 0.622 J/g°C and 300. g of water at 25.0°C. We can use the equation:

q = mcΔT

where q is the heat, m is the mass, c is the specific heat, and ΔT is the change in temperature. Since the metal is losing heat, the heat gained by the water is equal to the heat lost by the metal. Therefore, we set up the equation:

mcΔT = mcΔT

0.622 J/g°C * 71.0 g * (T - 25.0°C) = 4.184 J/g°C * 300. g * (25.0°C - T)

Solving this equation for T, the final temperature of the water, we get T = 9.0°C.