Respuesta :
A 295-g aluminum engine part at an initial temperature of 13.00°C absorbs 75.0 kJ of heat. The final temperature of the part is 295.5 °C.
The heat required, q, is equal to the Specific Heat Capacity, C, times the amount of the substance times the temperature change, that is (T final - T initial) in degrees C or K.
(Remember, a difference in temperature on the C scale is equivalent to that same amount on the Kelvin scale.)
So we have, q = C x m x (T final - T initial).
Let T final = T
Now, (T final - T initial) = (T) - (13 )
So,
q = C m (T final - T initial)
75kJ = (0.900 J/gK) (295g) [(T) - (13 ) ]
1kJ = 1000 J
75000 = (0.900 J/gK) (295g) [(T) - (13 ) ]
[(T) - (13 ) ] = 75000/ (0.900 x 295)
[(T) - (13 ) ] = 282.5
T = 282.5 +13
T = 295.5 °C
To learn more about the specific heat capacity please click on the link https://brainly.com/question/16559442
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