Calculate how much heat is required to bring 113 g of water from a temperature of 26.3°C to a temperature of 100°C. The specific heat of water is 4186 J/kg∙°C.
Q = m x c x delta t
m = 113 g
c = 4186 J/Kg C = 4.18 J/g C
delta t = 100 - 26.3 = 73.7 degree Celsius
Q = 113 X 4.18 X 73.7
Q = 34811.458 Joules = 3.48 Kilo Joules
Please do not forget to best the solution.
Content will be erased after question is completed.
Enter the email address associated with your account, and we will email you a link to reset your password.
Forgot your password?