Question

1. a 0.500 kg bar of a metal at 187°c was dropped into 500 g of water at 35°c. the final temperature of water was 45.9°c. assuming that no heat was lost to the surroundings, what is the specific heat of the metal? heat = mass × specific heat × δt where δt = t_final - t_initial 0.323 j/g°c 0.298 j/g°c 0.244 j/g°c 4.10 j/g°c

Explanation:

Step1: Convert metal mass to grams

$0.500\ \text{kg} = 0.500 \times 1000 = 500\ \text{g}$

Step2: Calculate heat gained by water

Mass of water $m_w=500\ \text{g}$, specific heat of water $c_w=4.184\ \text{J/g}^\circ\text{C}$, $\Delta t_w = 45.9^\circ\text{C} - 35^\circ\text{C}=10.9^\circ\text{C}$
$q_w = m_w \times c_w \times \Delta t_w = 500 \times 4.184 \times 10.9$
$q_w = 500 \times 45.6056 = 22802.8\ \text{J}$

Step3: Set heat lost by metal equal to $q_w$

Heat lost by metal $q_m = q_w = 22802.8\ \text{J}$, $\Delta t_m = 187^\circ\text{C} - 45.9^\circ\text{C}=141.1^\circ\text{C}$, $m_m=500\ \text{g}$
Rearrange $q = m \times c \times \Delta t$ to solve for $c_m$:
$c_m = \frac{q_m}{m_m \times \Delta t_m}$

Step4: Compute specific heat of metal

$c_m = \frac{22802.8}{500 \times 141.1} = \frac{22802.8}{70550} \approx 0.323\ \text{J/g}^\circ\text{C}$

Answer:

0.323 J/g°C