Question

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

Answer 1

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

Further explanation

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

Q in = Q out

Heat can be calculated using the formula:

Q = mc∆T

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:

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

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

cm = 509.18 J / Kg.K

Learn more

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Keywords: heat, temperature, thermal equilibrium

Answer 2

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 probably steel.