Answer:
Explanation:
Given
Force 
time interval 
Impulse is given by 
For two significant Figure

Answer:
Q = 1057.5 [cal]
Explanation:
In order to solve this problem, we must use the following equation of thermal energy.

where:
Q = heat energy [cal]
Cp = specific heat = 0.47 [cal/g*°C]
T_final = final temperature = 32 [°C]
T_initial = initial temperature = 27 [°C]
m = mass of the substance = 450 [g]
Now replacing:
![Q=450*0.47*(32-27)\\Q=1057.5[cal]](https://tex.z-dn.net/?f=Q%3D450%2A0.47%2A%2832-27%29%5C%5CQ%3D1057.5%5Bcal%5D)
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