Question

1. how much heat is needed to warm 0.054 kg of gold from 8.5°c to 99.0°c? note: gold has a specific heat of 129 j/kg·°c.
2. a 9.3 kg outdoor copper sculpture heats up during the day from 24°c to 77°c. how much energy was absorbed? note: copper has a specific heat of 385 j/kg·°c.
3. challenge: if it takes 820 joules of heat to warm a sample of zinc from 5.0°c to 55.0°c, what would be the mass of the zinc? note: zinc has a specific heat of 387 j/kg·°c.

Explanation:

Step1: Calculate the temperature change

For problem 5: $\Delta T = T_{final}-T_{initial}=99.0 - 8.5=90.5^{\circ}C$
For problem 6: $\Delta T = T_{final}-T_{initial}=77 - 24 = 53^{\circ}C$
For problem 7: $\Delta T = T_{final}-T_{initial}=55.0 - 5.0=50.0^{\circ}C$

Step2: Use the heat formula $Q = mc\Delta T$

Problem 5:

Given $m = 0.054\space kg$, $c = 129\space J/kg\cdot^{\circ}C$, $\Delta T=90.5^{\circ}C$
$Q=mc\Delta T=(0.054)(129)(90.5)$
$Q = 0.054\times129\times90.5=640.131\space J$

Problem 6:

Given $m = 9.3\space kg$, $c = 385\space J/kg\cdot^{\circ}C$, $\Delta T = 53^{\circ}C$
$Q=mc\Delta T=(9.3)(385)(53)$
$Q=9.3\times385\times53 = 9.3\times20405=190, 066.5\space J$

Problem 7:

Given $Q = 820\space J$, $c = 387\space J/kg\cdot^{\circ}C$, $\Delta T=50.0^{\circ}C$
From $Q = mc\Delta T$, we can solve for $m$: $m=\frac{Q}{c\Delta T}$
$m=\frac{820}{387\times50.0}=\frac{820}{19350}\approx0.0424\space kg$

Answer:

• Problem 5: $640.131\space J$
• Problem 6: $190066.5\space J$
• Problem 7: $0.0424\space kg$