Calculate the amount of heat given off by 640 g of water cooling from 76 °C to 28° C. Specific heat of water = 4.816 J/g C. Show your step by step process on how you have arrived at your answer. *

47947.52 J.

Explanation:

From the question,

Amount of heat given of (Q) = mc(t₁–t₂) Equation 1

Where m = mass of water, c = specific heat capacity of water, t₁ = initial temperature, t₂ = final temperature.

Given, m = 640 g = 640 g, c = 4.816 J/g°C, t₁ = 76 °C, t₂ = 28 °c.

Substitute these values into equation 1 above

Q = 640×4.816(48)

Q = 147947.52 J.

Hence the amount of heat given off is 47947.52 J.

explanation:

could i give you a exaplo that could you?

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm