Answer:
48.7°C
Explanation:
Step one:
given data
Quantity of heat = 4000J
mass of sample= 70g= 0.07kg
initial temperature T1=35°C
Water has a specific heat capacity of 4182 J/kg°C
Required
The final temperature T2
Step two:
Q= mcΔT
4000= 0.07*4182(T2-35)
4000=292.74(T2-35)
4000=292.74T2-10245.9
collect like terms
4000+10245.9=292.74T2
14245.9= 292.74T2
divide both sides by 292.74
T2= 14245.9/ 292.74
T2=48.7°C
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D all of the above Hope this helps
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