Question

A 0.5 kg piece of ice at -10 degrees Celsius is placed in 3 kg of water at 20 degrees Celsius. What temperature will the mixture finally be at? Consider that: cA=4186 J/kg°C and cH=2100 J/kg°C

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Answer to a math question A 0.5 kg piece of ice at -10 degrees Celsius is placed in 3 kg of water at 20 degrees Celsius. What temperature will the mixture finally be at? Consider that: cA=4186 J/kg°C and cH=2100 J/kg°C

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Corbin
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103 Answers
Step 1: Calculate the heat required to raise the temperature of ice from -10°C to 0°C.

The heat (Q) required to raise the temperature of the ice can be calculated using the formula:

Q = mcΔT

Given: mass of ice (m) = 0.5 kg, specific heat of ice (cA) = 2100 J/kg°C, initial temperature (Ti) = -10°C, final temperature (Tf) = 0°C

Q = 0.5 \times 2100 \times (0 - (-10))
Q = 0.5 \times 2100 \times 10 = 10500 J

Step 2: Calculate the heat required to melt the ice at 0°C.

The heat (Q) required to melt the ice can be calculated using the formula:

Q = mL
step 2.1:calculate the heat required to raise water(from ice) to final temperature
Q=0.5*4186*x


Given: latent heat of fusion of ice (L) = 334000 J/kg

Q = 0.5 \times 334000 = 167000 J

Step 3: Calculate the heat required to raise the temperature of the water from 20°C to the final temperature.

The heat (Q) required to raise the temperature of water can be calculated using the formula:

Q = mcΔT

Given: mass of water (m) = 3 kg, specific heat of water (cH) = 4186 J/kg°C, initial temperature (Ti) = 20°C, final temperature (Tf) = x°C

Q = 3 \times 4186 \times (x - 20)

Step 4: The total heat added to the system must be equal to zero for the final temperature to be reached.

10500+167000+0.5\times4186\times x+3\times4186\times(x-20)=0

Solving the equation gives:


177500+14651x-251160=0



14651x=73660

x ≈ 5°C

Therefore, the final temperature of the mixture will be approximately 5°C.

\textbf{Answer:} The final temperature of the mixture will be approximately 5°C.

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