Question

A 2.30-gram ice cube at water’s melting point is dropped into a glass of warm juice and gains enough joules of heat to melt completely. If the heat of fusion is 334 joules/gram, how much energy was absorbed?

Answer 1

The correct answer is

768

:)

Answer 2

Answer:- 768 J.

Solution:- 2.30 g of ice is present at water's melting point that is 0 degree C. Heat of fusion is given as 334 J per g and we are asked to calculate the energy absorbed to melt the ice.

334 J per g means 334 J of heat is required to melt one gram of ice. How much of heat would be required to melt 2.30 g of ice.

the equation used is,
`q=m.\Delta H_f_u_s`

where q is heat energy, m is mass and
`\Delta H_f_u_s` is the enthalpy of fusion.

Let's plug in the values:

`q=2.30g((334J)/(g))`

q = 768.2 J

If rounded for correct number of sig figs then it is 768 J. So, 768 J of energy is absorbed to melt 2.30 g of ice.