A machine part rotates at an angular speed of 0.76 rad/s; its speed is then increased to 2.96 rad/s using an angular acceleratio
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<h2>Δt = 71.67 °C</h2>
The temperature change of water is equal to 71.67 °C
<h3>Explanation:</h3>
Given:
Amount of transferred energy = 30,000 K J
Mass of water = 100 ml
Initial temperature = 20°C
To find the change in temperature of water.
Formula for Heat capacity is given by
Q = m×c×Δt ........................................(1)
where:
Q = Heat capacity of the substance (in J)
m=mass of the substance being heated in grams(g)
c = the specific heat of the substance in J/(g.°C)
Δt = Change in temperature (in °C)
Δt = (Final temperature - Initial temperature) = T(f) - T(i)
Q = 30,000 J
Mass of water = m = 100 ml
1 ml = 1 g ................................................(2)
Therefore m = 100 ml = 100 g
Specific heat of water is c = 4.186 J /g.
Δt = ?
Substituting these in equation (1), we get
Q = m×c×Δt
Rearranging the terms for Δt,
Δt =
Δt =
Δt = 71.67 °C