Calculate the temperature change in the water upon complete melting of the ice. Hint: Determine how much heat is absorbed by the melting ice and then use q =mC*Delta T to calculate the temperature change of the 314 g of water.

under the following conditions:
A 14.7 g ice cube is placed into 314 g of water.

can someone help me?? im completely lost. or can someone help me start the problem? thanks

there is no starting temperature or no temperature for the 314 g of water. that's why im lost. can someone help?

Sure! I can definitely help you with this problem. Let's break it down step by step.

Step 1: Calculate the heat absorbed by the melting ice.
When the ice cube melts, it absorbs heat from the surrounding water. The amount of heat absorbed can be calculated using the formula:

q = m * ΔHf

where q is the heat absorbed, m is the mass of the ice, and ΔHf is the heat of fusion of ice. The heat of fusion of ice is the amount of heat required to convert one gram of ice at 0°C into water at 0°C, which is 334 J/g.

So, in this case, with a mass of ice cube (m) of 14.7 g, the heat absorbed by the melting ice would be:

q = 14.7 g * 334 J/g

Step 2: Calculate the temperature change of the water.
Next, we can use the specific heat capacity of water to calculate the temperature change of the surrounding water. The specific heat capacity of water (C) is 4.186 J/g°C.

Using the formula: q = m * C * ΔT, we can rearrange it to find ΔT:

ΔT = q / (m * C)

In this case, the mass of water (m) is given as 314 g. Thus, we can substitute the known values into the equation:

ΔT = (14.7 g * 334 J/g) / (314 g * 4.186 J/g°C)

Step 3: Calculate the final answer.
Now, perform the calculations to find the temperature change (ΔT). The units of grams (g) will cancel out when dividing, leaving the units of Celsius (°C).

This will give you the temperature change of the water upon complete melting of the ice.