Marie1969
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A 40 g sample of water at an inital temperature of 40°C is heated to 99°C. How much heat did the water absorb? The specific heat of water is 4.18J/gºC


a. 413.82j
b.-9864j
c.9864.8j
d.-413.82j
​

Respuesta :

SvetkaChem

Answer:

C.

Explanation:

Q = mcΔt°

Q = 40g * 4.18 J/g*°C *(99°C - 40°C) = 9864.8 J

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