1. [tex]-8.78 \cdot 10^5 J[/tex]
The energy lost by the water is given by:
[tex]Q=m C_s \Delta T[/tex]
where
m = 3.0 kg = 3000 g is the mass of water
Cs = 4.179 J/g•°C is the specific heat
[tex]\Delta T=10.0C-80.0C=-70.0 C[/tex] is the change in temperature
Substituting,
[tex]Q=(3000 g)(4.179 J/gC)(-70.0 C)=-8.78 \cdot 10^5 J[/tex]
2. [tex]3.24 \cdot 10^4 J[/tex]
The energy added to the aluminium is given by:
[tex]Q=m C_s \Delta T[/tex]
where
m = 0.30 kg = 300 g is the mass of aluminium
Cs = 0.900 J/g•°C is the specific heat
[tex]\Delta T=150.0 C-30.0C =120.0 C[/tex] is the change in temperature
Substituting,
[tex]Q=(300 g)(0.900 J/gC)(120.0 C)=3.24 \cdot 10^4 J[/tex]
3a. [tex]-5.6^{\circ}C[/tex]
The temperature change of the water is given by
[tex]\Delta T=\frac{Q}{m C_s}[/tex]
where
[tex]Q = -232 kJ=-2.32\cdot 10^5 J[/tex] is the heat lost by the water
[tex]m=10.0 kg=10000 g[/tex] is the mass of water
Cs = 4.179 J/g•°C is the specific heat
Substituting,
[tex]\Delta T=\frac{-2.32\cdot 10^5 J}{(10000g)(4.179 J/gC)}=5.6^{\circ}C[/tex]
3b. [tex]+10.2^{\circ}C[/tex]
The temperature change of the copper is given by
[tex]\Delta T=\frac{Q}{m C_s}[/tex]
where
[tex]Q = 1.96 kJ=1960[/tex] is the heat added to the copper
[tex]m= 500 g[/tex] is the mass of copper
Cs = 0.385 J/g•°C is the specific heat
Substituting,
[tex]\Delta T=\frac{1960 J}{(500g)(0.385 J/gC)}=10.2^{\circ}C[/tex]
4. 42.9 g
The mass of the water sample is given by
[tex]m=\frac{Q}{C_S \Delta T}[/tex]
where
[tex]Q=4300 J[/tex] is the heat added
[tex]\Delta T=39 C-15 C=24C[/tex] is the temperature change
Cs = 4.179 J/g•°C is the specific heat
Substituting,
[tex]m=\frac{4300 J}{(4.179 J/gC)(24 C)}=42.9 g[/tex]
5. 115.5 J
The heat used to heat the copper is given by:
[tex]Q=m C_s \Delta T[/tex]
where
m = 5.0 g is the mass of copper
Cs = 0.385 J/g•°C is the specific heat
[tex]\Delta T=80.0 C-20.0C =60.0 C[/tex] is the change in temperature
Substituting,
[tex]Q=(5.0 g)(0.385 J/gC)(60.0 C)=115.5 J[/tex]
6. 0.185 J/g•°C
The specific heat of iron is given by:
[tex]C_s = \frac{Q}{m \Delta T}[/tex]
where
Q = -47 J is the heat released by the iron
m = 10.0 g is the mass of iron
[tex]\Delta T=25.0-50.4 C=-25.4 C[/tex] is the change in temperature
Substituting,
[tex]C_s = \frac{-47 J}{(10.0 g)(-25.4 C)}=0.185 J/gC[/tex]
