# Answer to Question #19697 in Inorganic Chemistry for Farhat

Question #19697

A 35.0 g block of copper at 350.0 degrees celcius is added to a beaker containing 525 g of water at 35.0 degrees celcius. What is the final temperature of water?

Expert's answer

Let us denote

m

c

t

m

c

t

T - final temperature.

we can write

m

Hence

T = (m

m

_{c}- mass of copper = 35 gc

_{c}- specific heat capacity of copper = 385 J/(kg*K)t

_{c}- temperature of copper = 350 C = 623 Km

_{w}- mass of water = 525gc

_{w}- specific heat capacity of water = 4200 J/(kg*K)t

_{w}- temperature of water = 35 C = 308 KT - final temperature.

we can write

m

_{c}*c_{c}*(t_{c}-T) = m_{w}*c_{w}*(T-t_{w})Hence

T = (m

_{c}*t_{c}*c_{c}+ m_{w}*t_{w}*c_{w}) / (m_{w}*c_{w}+ m_{c}*t_{c}) = 310K = 37 degrees celciusNeed a fast expert's response?

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