Question

Use the table below to answer the following questions. Substance Specific Heat (J/g•°C) water 4.179 aluminum 0.900 copper 0.385 iron 0.450 granite 0.790 1. When 3.0 kg of water is cooled from 80.0C to 10.0C, how much heat energy is lost? 2. How much heat is needed to raise a 0.30 kg piece of aluminum from 30.C to 150C? 3. Calculate the temperature change when: a) 10.0 kg of water loses 232 kJ of heat. b) 1.96 kJ of heat are added to 500. g of copper. 4. When heated, the temperature of a water sample increased from 15°C to 39°C. It absorbed 4300 joules of heat. What is the mass of the sample? 5. 5.0 g of copper was heated from 20°C to 80°C. How much energy was used to heat Cu? 6. The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 47 Joules of heat. What is the specific heat of iron? 7. The temperature of a sample of water increases

Answer 1

1. $-8.78 \cdot 10^5 J$

The energy lost by the water is given by:

$Q=m C_s \Delta T$

where

m = 3.0 kg = 3000 g is the mass of water

Cs = 4.179 J/g•°C is the specific heat

$\Delta T=10.0C-80.0C=-70.0 C$ is the change in temperature

Substituting,

$Q=(3000 g)(4.179 J/gC)(-70.0 C)=-8.78 \cdot 10^5 J$

2. $3.24 \cdot 10^4 J$

The energy added to the aluminium is given by:

$Q=m C_s \Delta T$

where

m = 0.30 kg = 300 g is the mass of aluminium

Cs = 0.900 J/g•°C is the specific heat

$\Delta T=150.0 C-30.0C =120.0 C$ is the change in temperature

Substituting,

$Q=(300 g)(0.900 J/gC)(120.0 C)=3.24 \cdot 10^4 J$

3a. $-5.6^{\circ}C$

The temperature change of the water is given by

$\Delta T=\frac{Q}{m C_s}$

where

$Q = -232 kJ=-2.32\cdot 10^5 J$ is the heat lost by the water

$m=10.0 kg=10000 g$ is the mass of water

Cs = 4.179 J/g•°C is the specific heat

Substituting,

$\Delta T=\frac{-2.32\cdot 10^5 J}{(10000g)(4.179 J/gC)}=5.6^{\circ}C$

3b. $+10.2^{\circ}C$

The temperature change of the copper is given by

$\Delta T=\frac{Q}{m C_s}$

where

$Q = 1.96 kJ=1960$ is the heat added to the copper

$m= 500 g$ is the mass of copper

Cs = 0.385 J/g•°C is the specific heat

Substituting,

$\Delta T=\frac{1960 J}{(500g)(0.385 J/gC)}=10.2^{\circ}C$

4. 42.9 g

The mass of the water sample is given by

$m=\frac{Q}{C_S \Delta T}$

where

$Q=4300 J$ is the heat added

$\Delta T=39 C-15 C=24C$ is the temperature change

Cs = 4.179 J/g•°C is the specific heat

Substituting,

$m=\frac{4300 J}{(4.179 J/gC)(24 C)}=42.9 g$

5. 115.5 J

The heat used to heat the copper is given by:

$Q=m C_s \Delta T$

where

m = 5.0 g is the mass of copper

Cs = 0.385 J/g•°C is the specific heat

$\Delta T=80.0 C-20.0C =60.0 C$ is the change in temperature

Substituting,

$Q=(5.0 g)(0.385 J/gC)(60.0 C)=115.5 J$

6. 0.185 J/g•°C

The specific heat of iron is given by:

$C_s = \frac{Q}{m \Delta T}$

where

Q = -47 J is the heat released by the iron

m = 10.0 g is the mass of iron

$\Delta T=25.0-50.4 C=-25.4 C$ is the change in temperature

Substituting,

$C_s = \frac{-47 J}{(10.0 g)(-25.4 C)}=0.185 J/gC$