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# 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

Corbin
4.6
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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