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} - mass of copper = 35 g

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

t_{c} - temperature of copper = 350 C = 623 K

m_{w} - mass of water = 525g

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

t_{w} - temperature of water = 35 C = 308 K

T - 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 celcius

m

c

t

m

c

t

T - final temperature.

we can write

m

Hence

T = (m

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