You take 323 g of a solid (melting point = 56.4°C, heat of fusion = 343 J/g) and let it melt in 758 g of water. The water temperature decreases from its initial temperature to 56.4°C. Calculate the initial temperature of the water.

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat required to melt the solid:

q = mΔHf
q = (323 g) (343 J/g)
q = 1.108 x 10^5 J

Next, we need to use the formula:

q = mCΔT

where q is the amount of heat transferred, m is the mass of the water, C is the specific heat of water (4.18 J/g°C), and ΔT is the change in temperature of the water.

Since the solid melts at its melting point and there is no further increase in temperature, we know that the amount of heat lost by the solid (q) is equal to the amount of heat gained by the water (q). Thus:

1.108 x 10^5 J = (758 g) (4.18 J/g°C) (Ti - 56.4°C)

Solving for Ti, we get:

Ti = 91.4°C

Therefore, the initial temperature of the water was 91.4°C.

You Take 323 G of a Solid (Melting Point =