Skip to main content
Ch.5 - Thermochemistry
Chapter 5, Problem 106b

A coffee-cup calorimeter of the type shown in Figure 5.18 contains 150.0 g of water at 25.1°C A 121.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g-K The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1°C (b) Determine the amount of heat gained by the water. The specific heat of water is 4.184 J/1gK.

Verified step by step guidance
1
Identify the known values: mass of water (m_w) = 150.0 g, initial temperature of water (T_i,w) = 25.1°C, final temperature of water (T_f) = 30.1°C, specific heat of water (c_w) = 4.184 J/g-K.
Calculate the change in temperature for the water: ΔT_w = T_f - T_i,w.
Use the formula for heat gained or lost: q = m * c * ΔT, where q is the heat gained or lost, m is the mass, c is the specific heat, and ΔT is the change in temperature.
Substitute the known values into the formula: q_w = m_w * c_w * ΔT_w.
Solve for q_w to find the amount of heat gained by the water.

Verified Solution

Video duration:
1m
This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin). It is a material-specific property that varies between different substances. For example, water has a high specific heat capacity of 4.184 J/g-K, meaning it can absorb a lot of heat without a significant change in temperature, which is crucial in calorimetry calculations.
Recommended video:
Guided course
02:19
Heat Capacity

Calorimetry

Calorimetry is the science of measuring the heat of chemical reactions or physical changes. In a calorimeter, the heat lost by a hot object (like the copper block) is equal to the heat gained by the cooler substance (like water). This principle allows us to calculate the heat transfer in a system, which is essential for solving problems involving temperature changes and specific heat.
Recommended video:
Guided course
00:50
Constant-Volume Calorimetry

Heat Transfer Equation

The heat transfer equation, often expressed as q = mcΔT, relates the heat gained or lost (q) to the mass (m), specific heat capacity (c), and the change in temperature (ΔT) of a substance. In this context, it allows us to calculate the amount of heat gained by the water when the copper block is added, by substituting the known values of mass, specific heat, and temperature change into the equation.
Recommended video:
Guided course
02:19
Heat Capacity
Related Practice
Open Question
Both oxyhydrogen torches and fuel cells use the following reaction to produce energy: 2 H2(g) + O2(g) → 2 H2O(l). Both processes occur at constant pressure. In both cases, the change in state of the system is the same: the reactant is oxyhydrogen (“Knallgas”) and the product is water. Yet, with an oxyhydrogen torch, the heat evolved is large, and with a fuel cell, it is small. If heat at constant pressure is considered to be a state function, why does it depend on path?
Textbook Question

A house is designed to have passive solar energy features. Brickwork incorporated into the interior of the house acts as a heat absorber. Each brick weighs approximately 1.8 kg. The specific heat of the brick is 0.85 J/g•K. How many bricks must be incorporated into the interior of the house to provide the same total heat capacity as 1.7⨉103 gal of water?

1818
views
Textbook Question

A coffee-cup calorimeter of the type shown in Figure 5.18 contains 150.0 g of water at 25.1°C A 121.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g-K The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1°C. (a) Determine the amount of heat, in J, lost by the copper block.

808
views
Textbook Question

A coffee-cup calorimeter of the type shown in Figure 5.18 contains 150.0 g of water at 25.1°C A 121.0-g block of copper metal is heated to 100.4°C by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g-K The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1°C (d) What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter?

1200
views
1
comments
Textbook Question

(b) Assuming that there is an uncertainty of 0.002 °C in each temperature reading and that the masses of samples are measured to 0.001 g, what is the estimated uncertainty in the value calculated for the heat of combustion per mole of caffeine?

911
views
1
rank
Open Question
The corrosion (rusting) of iron in oxygen-free water includes the formation of iron(II) hydroxide from iron by the following reaction: Fe(s) + 2 H2O(l) → Fe(OH)2(s) + H2(g). (b) Calculate the number of grams of Fe needed to release enough energy to increase the temperature of 250 mL of water from 22 to 30 °C.