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 [tex]c[/tex].
Energy Exchange = Specific Heat ⨯ Mass ⨯ Temperature Change
Energy the Hot Lump Lost = Energy the Cold Water Gained
Water has a specific heat of [tex]4.2 \; \text{J} \cdot \text{g} ^{-1}\cdot \text{K}^{-1}[/tex] and a density of [tex]1 \; \text{g} \cdot \text{ml}^{-1}[/tex]. 25.0 milliliters of water thus has a mass of 25.0 grams.
[tex]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}[/tex]
Steel has a specific heat of approximately [tex]0.51 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}[/tex]. This substance is thus probably steel.
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
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:
[tex] \displaystyle m_mc_m (T_m-T) = m_wc_w (T-Tw) [/tex]
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:
[tex] \displaystyle m_mc_m (T_m-T) = m_wc_w (T-Tw) [/tex]
[tex] \displaystyle 0.1157.c_m (441.3-349.5) = 0.025.4200 (349.5-298) [/tex]
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Keywords: heat, temperature, thermal equilibrium
