Question

find the specific heat of the metal.
2 water temperature change
50 g of water starts at 20°c and absorbs 4,180 j of heat.
(specific heat of water = 4.18 j/g·°c)
what is the final temperature of the water?
3 thermal equilibrium between metal and water
a 100 g metal at 80°c is placed in 150 g of water at 20°c.
the final temperature is 25°c.
(specific heat of water = 4.18 j/g·°c)
find the specific heat of the metal.
4 melting ice in water
30 g of ice at 0°c is added to 200 g of water at 25°c.
all ice melts and the final temperature is 10°c.
(specific heat of water = 4.18 j/g·°c)
calculate the heat lost by the water.

Explanation:

(Problem 2):

Step1: Recall heat transfer formula

The formula for heat transfer is $Q = mc\Delta T = mc(T_f - T_i)$, where $Q$ is heat, $m$ is mass, $c$ is specific heat, $T_f$ is final temp, $T_i$ is initial temp.

Step2: Rearrange for final temperature

$T_f = T_i + \frac{Q}{mc}$

Step3: Substitute given values

$m=50\ \text{g}$, $c=4.18\ \text{J/g·°C}$, $T_i=20^\circ\text{C}$, $Q=4180\ \text{J}$
$\frac{Q}{mc} = \frac{4180}{50 \times 4.18} = 20^\circ\text{C}$
$T_f = 20 + 20 = 40^\circ\text{C}$

(Problem 3):

Step1: Set heat lost = heat gained

Heat lost by metal: $Q_m = m_m c_m (T_{i,m} - T_f)$; Heat gained by water: $Q_w = m_w c_w (T_f - T_{i,w})$
$m_m c_m (T_{i,m} - T_f) = m_w c_w (T_f - T_{i,w})$

Step2: Rearrange for metal specific heat

$c_m = \frac{m_w c_w (T_f - T_{i,w})}{m_m (T_{i,m} - T_f)}$

Step3: Substitute given values

$m_m=100\ \text{g}$, $T_{i,m}=80^\circ\text{C}$, $m_w=150\ \text{g}$, $T_{i,w}=20^\circ\text{C}$, $T_f=25^\circ\text{C}$, $c_w=4.18\ \text{J/g·°C}$
$c_m = \frac{150 \times 4.18 \times (25-20)}{100 \times (80-25)} = \frac{150 \times 4.18 \times 5}{100 \times 55} = 0.576\ \text{J/g·°C}$

(Problem 4):

Step1: Use heat transfer formula

Heat lost by water: $Q = mc(T_i - T_f)$ (since heat is lost, $T_i > T_f$)

Step2: Substitute given values

$m=200\ \text{g}$, $c=4.18\ \text{J/g·°C}$, $T_i=25^\circ\text{C}$, $T_f=10^\circ\text{C}$
$Q = 200 \times 4.18 \times (25-10) = 200 \times 4.18 \times 15 = 12540\ \text{J}$

Answer:

1. (Problem 2) Final temperature of water: $40^\circ\text{C}$
2. (Problem 3) Specific heat of the metal: $0.576\ \text{J/g·°C}$
3. (Problem 4) Heat lost by the water: $12540\ \text{J}$