The same amount of heat entering identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.40 kcal of heat enters 1.83 kg of the following, originally at 28.2°C. The specific heat capacity for each material is given in square brackets below. (a) water [1.00 kcal/(kg · °C)] °C (b) concrete [0.20 kcal/(kg · °C)] °C (c) steel [0.108 kcal/(kg · °C)] °C (d) mercury [0.0333 kcal/(kg · °C)] °C
The same amount of heat entering identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.40 kcal of heat enters 1.83 kg of the following, originally at 28.2°C. The specific heat capacity for each material is given in square brackets below. (a) water [1.00 kcal/(kg · °C)] °C (b) concrete [0.20 kcal/(kg · °C)] °C (c) steel [0.108 kcal/(kg · °C)] °C (d) mercury [0.0333 kcal/(kg · °C)] °C
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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![**Heat and Temperature Change in Different Substances**
In this exercise, we explore how the same amount of heat entering identical masses of different substances affects their temperature changes. Specifically, calculate the final temperature when 1.40 kcal of heat enters 1.83 kg of different materials that originally have a temperature of 28.2°C. The specific heat capacity for each material is as follows:
- **Water**: `[1.00 kcal/(kg·°C)]`
- **Concrete**: `[0.20 kcal/(kg·°C)]`
- **Steel**: `[0.108 kcal/(kg·°C)]`
- **Mercury**: `[0.0333 kcal/(kg·°C)]`
Use the formula for temperature change:
\[ \Delta T = \frac{Q}{m \cdot c} \]
Where:
- \( \Delta T \) is the change in temperature
- \( Q \) is the heat energy added (1.40 kcal)
- \( m \) is the mass of the substance (1.83 kg)
- \( c \) is the specific heat capacity
After finding \( \Delta T \), add it to the original temperature (28.2°C) to find the final temperature.
### Calculations for Each Substance
(a) **Water** `[1.00 kcal/(kg·°C)]`:
\[ \frac{1.40 \text{ kcal}}{1.83 \text{ kg} \cdot 1.00 \text{ kcal/(kg·°C)}} = \Delta T_{water} \]
(b) **Concrete** `[0.20 kcal/(kg·°C)]`:
\[ \frac{1.40 \text{ kcal}}{1.83 \text{ kg} \cdot 0.20 \text{ kcal/(kg·°C)}} = \Delta T_{concrete} \]
(c) **Steel** `[0.108 kcal/(kg·°C)]`:
\[ \frac{1.40 \text{ kcal}}{1.83 \text{ kg} \cdot 0.108 \text{ kcal/(kg·°C)}} = \Delta T_{steel} \]
(d) **Mercury** `[0.0333 kcal/(kg·°C)]`:
\[ \frac{1.40 \text{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5362cbea-9427-4b2b-99cb-57a208edd853%2Fe2681664-8656-4fec-a44a-5355c736b3c2%2Fnvbpc2w.png&w=3840&q=75)
Transcribed Image Text:**Heat and Temperature Change in Different Substances**
In this exercise, we explore how the same amount of heat entering identical masses of different substances affects their temperature changes. Specifically, calculate the final temperature when 1.40 kcal of heat enters 1.83 kg of different materials that originally have a temperature of 28.2°C. The specific heat capacity for each material is as follows:
- **Water**: `[1.00 kcal/(kg·°C)]`
- **Concrete**: `[0.20 kcal/(kg·°C)]`
- **Steel**: `[0.108 kcal/(kg·°C)]`
- **Mercury**: `[0.0333 kcal/(kg·°C)]`
Use the formula for temperature change:
\[ \Delta T = \frac{Q}{m \cdot c} \]
Where:
- \( \Delta T \) is the change in temperature
- \( Q \) is the heat energy added (1.40 kcal)
- \( m \) is the mass of the substance (1.83 kg)
- \( c \) is the specific heat capacity
After finding \( \Delta T \), add it to the original temperature (28.2°C) to find the final temperature.
### Calculations for Each Substance
(a) **Water** `[1.00 kcal/(kg·°C)]`:
\[ \frac{1.40 \text{ kcal}}{1.83 \text{ kg} \cdot 1.00 \text{ kcal/(kg·°C)}} = \Delta T_{water} \]
(b) **Concrete** `[0.20 kcal/(kg·°C)]`:
\[ \frac{1.40 \text{ kcal}}{1.83 \text{ kg} \cdot 0.20 \text{ kcal/(kg·°C)}} = \Delta T_{concrete} \]
(c) **Steel** `[0.108 kcal/(kg·°C)]`:
\[ \frac{1.40 \text{ kcal}}{1.83 \text{ kg} \cdot 0.108 \text{ kcal/(kg·°C)}} = \Delta T_{steel} \]
(d) **Mercury** `[0.0333 kcal/(kg·°C)]`:
\[ \frac{1.40 \text{
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