Question

silver pellets with a mass of 1.0 g and a temperature of 85 °c are added to 220 g of water at 14 °c. how many pellets must be added to increase the equilibrium temperature of the system to 25 °c? assume no heat is exchanged with the surroundings.

a. 7.2×10² pellets
b. 3.4×10⁵ pellets
c. 9.5×10¹ pellets
d. 8.3×10⁴ pellets
e. 1.1×10³ pellets

Explanation:

Step1: Recall heat - transfer formula

The heat transfer formula is $Q = mc\Delta T$, where $Q$ is the heat transferred, $m$ is the mass, $c$ is the specific - heat capacity, and $\Delta T$ is the change in temperature. The specific - heat capacity of water $c_w=4.186\ J/(g\cdot^{\circ}C)$ and the specific - heat capacity of silver $c_{Ag}=0.235\ J/(g\cdot^{\circ}C)$.

Step2: Calculate the heat gained by water

The mass of water $m_w = 220\ g$, the initial temperature of water $T_{w1}=14^{\circ}C$, and the final temperature $T = 25^{\circ}C$. So, $\Delta T_w=T - T_{w1}=25 - 14=11^{\circ}C$. Then the heat gained by water $Q_w=m_wc_w\Delta T_w$. Substituting the values, we get $Q_w = 220\times4.186\times11=220\times46.046 = 10130.12\ J$.

Step3: Calculate the heat lost by one silver pellet

The mass of one silver pellet $m_{Ag}=1.0\ g$, the initial temperature of silver $T_{Ag1}=85^{\circ}C$, and the final temperature $T = 25^{\circ}C$. So, $\Delta T_{Ag}=T_{Ag1}-T=85 - 25 = 60^{\circ}C$. Then the heat lost by one silver pellet $Q_{Ag}=m_{Ag}c_{Ag}\Delta T_{Ag}=1\times0.235\times60 = 14.1\ J$.

Step4: Calculate the number of silver pellets

Let the number of silver pellets be $n$. Since the heat lost by the silver pellets is equal to the heat gained by the water (assuming no heat exchange with the surroundings), we have $nQ_{Ag}=Q_w$. Then $n=\frac{Q_w}{Q_{Ag}}=\frac{10130.12}{14.1}\approx718.45\approx7.2\times10^{2}$.

Answer:

a. $7.2\times 10^{2}$ pellets