Question

200 g of water in a glass is at 10.0°C when an ice cube is placed in the glass. What is the temperature of the water after it has transferred 4,200 J of thermal energy to the ice cube? °C​

Answer 1

To find the temperature of the water after transferring thermal energy to the ice cube, we can use the equation Q = mcΔT. By calculating the change in temperature using the given thermal energy and the mass and specific heat capacity of water, we can determine the final temperature.

Step-by-step explanation:

The temperature of the water after it has transferred 4,200 J of thermal energy to the ice cube can be determined using the equation:

Q = mcΔT

Where Q is the thermal energy transferred, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

From the given information, we can calculate the change in temperature of the water as follows:

1. Calculate the heat capacity of the water using the equation: C = mc. Here, m is the mass of the water (200 g) and c is the specific heat capacity of water (4.184 J/(g•°C)). Therefore, C = 200 g * 4.184 J/(g•°C).
2. Determine the change in temperature using the equation: ΔT = Q/C. Here, Q is the thermal energy transferred (4,200 J) and C is the heat capacity of the water. Therefore, ΔT = 4,200 J / (200 g * 4.184 J/(g•°C)).
3. Substitute the values into the equation and solve for ΔT: ΔT = 4,200 J / (200 g * 4.184 J/(g•°C)).

Answer 2

20.0°C

Step-by-step explanation:

To find the final temperature of the water after it has transferred thermal energy to the ice cube, we need to use the equation:

Q = cmΔT

where Q is the amount of thermal energy transferred, c is the specific heat capacity of the substance, m is the mass of the substance, and ΔT is the change in temperature.

In this case, we are given that Q = 4,200 J and m = 200 g. The specific heat capacity of water is 4.184 J/g°C.

Substituting these values into the equation gives us:

4,200 J = (4.184 J/g°C)(200 g)(ΔT)

Solving for ΔT gives us:

ΔT = 4,200 J / [(4.184 J/g°C)(200 g)] = 10°C

Since the initial temperature of the water was 10.0°C, the final temperature of the water is 10.0°C + 10°C = 20.0°C.

Therefore, the temperature of the water in the glass is 20.0°C after it has transferred 4,200 J of thermal energy to the ice cube.