Will someone please help me with this Physics question.

Question

Using the following equation, 

| m_mC_m (T_f - T_i,m) |=| m_wC_w (T_f - T_i,w) |

A student mixes .075 kg of an unknown substance at 96.5°C with .075 kg of water at 25.0°C. If the final temperature of the system is 31.15°C, what is the specific heat capacity of the substance? Use 4184 J / kg °C for the specific heat capacity of water.

Answer 1

Given:

m_m = 0.075 kg

m_w = 0.075 kg

T_{i, m} = 96.5°C

T_{i, w} = 25.0°C

C_w = 4184 J / kg * °C

Find:

C_m = ? J / kg * °C

Solution:

So, absolute values are weird, but there's some possible algebra you can do to them.

| a * b | = | a | * | b | and additionally | a | = | a * b | / | b |

The restriction comes in when:

| a - b | ≠ | a | - | b |

Also, if a value is already positive, and consequently greater than zero, the absolute value can be satisfied since its argument is positive and the bars can be dropped

If and only if g > 0, then | g | = g

So:

| m_mC_m (T_f - T_i,m) |=| m_wC_w (T_f - T_i,w) |

| m_m | * | C_m | * | T_f - T_i,m | = | m_w | * | C_w | * | T_f - T_i,w |

| C_m | = C_m = [ m_w * C_w * | T_f - T_i,w | ] / [ m_m * | T_f - T_i,m | ]

C_m = (0.075 kg) * (4184 J / kg * °C) * | 31.15°C - 25.0°C | / [ (0.075 kg) * | 31.15°C - 96.5°C |]

∴ C_m ≈ 393.75 J / kg * °C

I hope this helps!

The interpretation of m_m is strange. I assume that even if you mix the water and the unknown that the unknown in the mixture still retains its characteristics and just matches temperature with its neighboring mixture of water meaning that m_m would be just the mass of the unknown substance. Even if m_m is for the mass of the mixture after, it would scale the answer by a half to ~196.88 J / kg * °C.

Either way, message me in the comments with any questions, comments, or concerns!

Answer 2

0.75kg·C_m(96.5°C - 31.15°C) = 0.75kg·4184J/kg°C(31.15°C - 25.0°C);

C_m = 393.75 J/kg°C