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
2 answers:
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
<h3>Q in = Q out </h3>Heat can be calculated using the formula:
<h3>Q = mc∆T </h3>An unknown substance is placed in 25.0 mL of water and there will be heat transfer:
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:
<h3> Learn more </h3>
the difference between temperature and heat
Specific heat
relationships among temperature, heat, and thermal energy.
When heat is added to a substance
Keywords: heat, temperature, thermal equilibrium
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 .
Energy Exchange = Specific Heat ⨯ Mass ⨯ Temperature Change
Energy the Hot Lump Lost = Energy the Cold Water Gained
Water has a specific heat of and a density of
. 25.0 milliliters of water thus has a mass of 25.0 grams.
Steel has a specific heat of approximately . This substance is thus <em>probably </em>steel.
You might be interested in
Answer:
q(problem 1) = 25,050 joules; q(problem 2) = 4.52 x 10⁶ joules
Explanation:
To understand these type problems one needs to go through a simple set of calculations relating to the 'HEATING CURVE OF WATER'. That is, consider the following problem ...
=> Calculate the total amount of heat needed to convert 10g ice at -10°C to steam at 110°C. Given are the following constants:
Heat of fusion (ΔHₓ) = 80 cal/gram
Heat of vaporization (ΔHv) = 540 cal/gram
specific heat of ice [c(i)] = 0.50 cal/gram·°C
specific heat of water [c(w)] = 1.00 cal/gram·°C
specific heat of steam [c(s)] = 0.48 cal/gram·°C
Now, the problem calculates the heat flow in each of five (5) phase transition regions based on the heating curve of water (see attached graph below this post) ... Note two types of regions (1) regions of increasing slopes use q = mcΔT and (2) regions of zero slopes use q = m·ΔH.
q(warming ice) = m·c(i)·ΔT = (10g)(0.50 cal/g°C)(10°C) = 50 cal
q(melting) = m·ΔHₓ = (10g)(80cal/g) 800 cal
q(warming water) = m·c(w)·ΔT = (10g)(1.00 cal/g°C)(100°C) = 1000 cal
q(evaporation of water) = m·ΔHv = (10g)(540cal/g) = 5400 cal
q(heating steam) = m·c(s)·ΔT = (10g)(0.48 cal/g°C)(10°C) = 48 cal
Q(total) = ∑q = (50 + 800 + 1000 + 5400 + 48) = 7298 cals. => to convert to joules, multiply by 4.184 j/cal => q = 7298 cals x 4.184 j/cal = 30,534 joules = 30.5 Kj.
Now, for the problems in your post ... they represent fragments of the above problem. All you need to do is decide if the problem contains a temperature change (use q = m·c·ΔT) or does NOT contain a temperature change (use q = m·ΔH).
Problem 1: Given Heat of Fusion of Water = 334 j/g, determine heat needed to melt 75g ice.
Since this is a phase transition (melting), NO temperature change occurs; use q = m·ΔHₓ = (75g)(334 j/g) = 25,050 joules.
Problem 2: Given Heat of Vaporization = 2260 j/g; determine the amount of heat needed to boil to vapor 2 Liters water ( = 2000 grams water ).
Since this is a phase transition (boiling = evaporation), NO temperature change occurs; use q = m·ΔHf = (2000g)(2260 j/g) = 4,520,000 joules = 4.52 x 10⁶ joules.
Problems containing a temperature change:
NOTE: A specific temperature change will be evident in the context of problems containing temperature change => use q = m·c·ΔT. Such is associated with the increasing slope regions of the heating curve. Good luck on your efforts. Doc :-)
