Question

Typically, water runs through the baseboard copper tubing and, therefore, fresh hot water is constantly running through the piping. However, consider a pipe where water was allowed to sit in the pipe. The hot water cools as it sits in the pipe. What is the temprature change, (ΔT), of the water if 199.0 g of water sat in the copper pipe from part A, releasing 2793 J of energy to the pipe? The specific heat of water is 4.184 (J/g)⋅∘C.

Answer 1

M = 199.0 g, the mass of water in the pipe.
c = 4.184 J/(g-C), the specific heat of water
Q = 2793 J, thermal energy released.

Use the formula
Q = m*c*ΔT
where
ΔT is the temperature change.

That is,
$(199.0 \, g)*(4.184 \, \frac{J}{g-^{o}C} )*(\Delta T \, ^{o}C) = (2793 \, J)\\\\ \Delta T = \frac{2793}{(199)(4.184)} =3.355 \, ^{o}C$

Answer: ΔT = 3.4 °C (nearest tenth)