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 180.5 g of water sat in the copper pipe from part a, releasing 2494 j of energy to the pipe? the specific heat of water is 4.184 (j/g)⋅∘c.

-

3.3° C

Explanation: -

Mass of water m = 180.5 g

Energy released as heat Q = 2494 J

Specific heat is defined as the heat required to raise the temperature of the unit mass of a given substance by 1 C.

Specific heat of water Cp = 4.184 (J/g)⋅∘C

Using the formula

Q = m x Cp x ΔT

We get temperature change ΔT = Q / (m x Cp)

= 2494 J / ( 180.5 g x 4.184 (J/g)⋅∘C

= 3.3° C

Thus the temprature change, (ΔT), of the wateris 3.3 °C if 180.5 g of water sat in the copper pipe from part A, releasing 2494 J of energy to the pipe

your answer is c) plasma

~*"auba14"*~