The initial temperature of the metal bolt is 80.8 °C
We'll begin by calculating the heat absorbed by the water.
- Mass of water (M) = 0.15 Kg
- Initial temperature (T₁) = 21 °C
- Final temperature (T₂) = 25 °C
- Change in temperature (ΔT) = T₂ – T₁ = 25 – 21 = 4 °C
- Specific heat capacity of water (C) = 4184 J/KgºC
Q = MCΔT
Q = 0.15 × 4184 × 4
Q = 2510.4 J
Finally, we shall determine the initial temperature of the metal bolt.
- Heat absorbed by water = 2510.4 J
- Heat released by metal (Q) = –2510.4 J
- Mass of metal (M) = 0.050 Kg
- Final temperature (T₂) = 25 °C
- Specific heat capacity of metal (C) = 899 J/Kg°C
- Initial temperature (T₁) =?
Q = MC(T₂ – T₁)
–2510.4 = 0.050 × 899 (25 – T₁)
–2510.4 = 44.95 (25 – T₁)
Clear bracket
–2510.4 = 1123.75 – 44.95T₁
Collect like terms
–2510.4 – 1123.75 = –44.95T₁
–3634.15 = –44.95T₁
Divide both side by –44.95
T₁ = –3634.15 / –44.95
T₁ = 80.8 °C
Thus, the initial temperature of the metal is 80.8 °C.
Learn more about heat tranfer:
brainly.com/question/26034272
Answer:
the angle is given by
Tan theta = 35/59 = 0.59
so theta = Tan ^-1 ( 0.59 )
theta = 30.54 deg.
The momentum change =mass*velocity change. But sincevelocity change is not known another strategy must be used to find the momentum change. The strategy involves first finding the impulse (F*t = 1.0 N*s). Since impulse = momentum change, the answer is 1.0 N*s.
