A 40 kg metal block at 600°C is dropped into 300 kg of oil at 25°C. If the final temperature is 40°C, what is the specific heat capacity of the metal? The specific heat capacity of oil is 2100 J/kg·°C

The heat lost by the metal is equal to the heat gained by the oil.
For the metal:
Mass, $m_m = 40 \text{ kg}$
Initial temperature, $T_{m,i} = 600^\circ\text{C}$
Final temperature, $T_{m,f} = 40^\circ\text{C}$
Specific heat capacity, $c_m = ?$

For the oil:
Mass, $m_o = 300 \text{ kg}$
Initial temperature, $T_{o,i} = 25^\circ\text{C}$
Final temperature, $T_{o,f} = 40^\circ\text{C}$
Specific heat capacity, $c_o = 2100 \text{ J/kg}\cdot^\circ\text{C}$

The principle is $Q_{lost, metal} = Q_{gained, oil}$.
The formula for heat transfer is $Q = mc\Delta T$.

For the metal:

$$\Delta T_m = T_{m,i} - T_{m,f} = 600^\circ\text{C} - 40^\circ\text{C} = 560^\circ\text{C}$$

For the oil:

$$\Delta T_o = T_{o,f} - T_{o,i} = 40^\circ\text{C} - 25^\circ\text{C} = 15^\circ\text{C}$$

$$m_m c_m \Delta T_m = m_o c_o \Delta T_o$$ $$(40 \text{ kg}) \cdot c_m \cdot (560^\circ\text{C}) = (300 \text{ kg}) \cdot (2100 \text{ J/kg}\cdot^\circ\text{C}) \cdot (15^\circ\text{C})$$ $$22400 \cdot c_m = 9450000$$ $$c_m = \frac{9450000}{22400}$$ $$c_m = 421.875 \text{ J/kg}\cdot^\circ\text{C}$$

The specific heat capacity of the metal is $\boxed{\text{421.875 J/kg}\cdot^\circ\text{C}}$.

13. How much water is frozen?

The heat gained by the ice to warm up to $0^\circ\text{C}$ is equal to the heat lost by the water as it freezes.
For the ice:
Mass, $m_{ice} = 100 \text{ g} = 0.1 \text{ kg}$
Initial temperature, $T_{ice,i} = -20^\circ\text{C}$
Final temperature, $T_{ice,f} = 0^\circ\text{C}$
Specific heat capacity of ice, $c_{ice} = 2302 \text{ J/kg}\cdot^\circ\text{C}$

For the water:
Initial temperature, $T_{water,i} = 0^\circ\text{C}$
Latent heat of fusion, $L_f = 3.35 \times 10^5 \text{ J/kg}$
Mass of water frozen, $m_{frozen} = ?$

The principle is $Q_{gained, ice} = Q_{lost, water}$.

$$Q_{gained, ice} = m_{ice} c_{ice} \Delta T_{ice}$$ $$\Delta T_{ice} = T_{ice,f} - T_{ice,i} = 0^\circ\text{C} - (-20^\circ\text{C}) = 20^\circ\text{C}$$ $$Q_{gained, ice} = (0.1 \text{ kg}) \cdot (2302 \text{ J/kg}\cdot^\circ\text{C}) \cdot (20^\circ\text{C})$$ $$Q_{gained, ice} = 4604 \text{ J}$$

The heat lost by the water as it freezes is given by $Q_{lost, water} = m_{frozen} L_f$.
Since $Q_{gained, ice} = Q_{lost, water}$:

$$4604 \text{ J} = m_{frozen} \cdot (3.35 \times 10^5 \text{ J/kg})$$ $$m_{frozen} = \frac{4604 \text{ J}}{3.35 \times 10^5 \text{ J/kg}}$$ $$m_{frozen} \approx 0.013743 \text{ kg}$$

To express this in grams:

$$m_{frozen} \approx 0.013743 \text{ kg} \times 1000 \text{ g/kg} \approx 13.743 \text{ g}$$

The mass of water frozen is approximately $\boxed{\text{0.0137 kg}}$.