Question

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
​

Answer 1

C.

Step-by-step explanation:

Q = mcΔt°

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