A hot lump of 115.7 g of an unknown substance initially at 168.3°C is placed in 25.0 mL of water initially at 25.0°C and allowed

Question

4 years ago

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A hot lump of 115.7 g of an unknown substance initially at 168.3°C is placed in 25.0 mL of water initially at 25.0°C and allowed

to reach thermal equilibrium. The final temperature of the system is 76.5°C. What is the identity of the unknown substance? Assume no heat is lost to the surroundings

Chemistry

2 answers:

Xelga [282] · Answer 1 · 2021-11-29T18:49:14+03:00

The specific heat capacity of the substance: 509.18 J / Kg.K, which approaches the specific heat capacity of metals table chart is <u>Steel, Mild </u>

<h3>Further explanation </h3>

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released

Heat can be calculated using the formula:

An unknown substance is placed in 25.0 mL of water and there will be heat transfer:

$\displaystyle m_mc_m (T_m-T) = m_wc_w (T-Tw)$

m = substance

w = water

T = the final temperature of the mixture

A hot lump of 115.7 g of an unknown substance initially at 168.3 ° C is placed in 25.0 mL of water initially at 25.0 ° C and allowed to reach thermal equilibrium. The final temperature of the system is 76.5 ° C

known:

m of substance = 115.7 g = 0.1157 kg

Tm = 168.3 ◦C + 273 = 441.3 K

m of water = 25 ml = 25 g = 0.025 kg

Tw = 25 ◦C + 273 = 298 K

cw = 4,200 J / kg C

T = 76.5 + 273 = 349.5 K

the specific heat capacity of the substance:

$\displaystyle m_mc_m (T_m-T) = m_wc_w (T-Tw)$

$\displaystyle 0.1157.c_m (441.3-349.5) = 0.025.4200 (349.5-298)$

<h3> Learn more </h3>

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Keywords: heat, temperature, thermal equilibrium

marshall27 [118] · Answer 2 · 2021-11-29T17:06:45+03:00

The substance has the specific heat capacity of steel and is therefore probably steel.

Let the specific heat of the unknown lump of substance be $c$ .

Energy Exchange = Specific Heat ⨯ Mass ⨯ Temperature Change

Energy the Hot Lump Lost = Energy the Cold Water Gained

Water has a specific heat of $4.2 \; \text{J} \cdot \text{g} ^{-1}\cdot \text{K}^{-1}$ and a density of $1 \; \text{g} \cdot \text{ml}^{-1}$ . 25.0 milliliters of water thus has a mass of 25.0 grams.

$115.7 \times (168.3 - 76.5) \; c = 4.2 \times 25.0 \times (76.5 - 25.0) \\c = 0.51 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}$

Steel has a specific heat of approximately $0.51 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}$ . This substance is thus <em>probably </em>steel.