Question

answer all of the questions to the best of your ability.
specific heat capacity for different materials
substance | specific heat (j/kg•°c) | substance | specific heat (j/kg•°c)
aluminum | 900 | mercury | 140
alcohol (ethyl) | 2400 | sand | 800
copper | 390 | silver | 230
glass | 840 | water | 4180
iron or steel | 450 | wood | 1700
lead | 130 | human body | 3470
marble | 860
part 1: energy transfer through heating

1. how much heat, in joules, is required to raise the temperature of 0.10 kg of aluminum from 5°c to 85°c?

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Explanation:

Step1: Recall the heat formula

The formula for heat energy \( Q \) is \( Q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature.

Step2: Identify values

Given \( m = 0.10 \, \text{kg} \), \( c_{\text{aluminum}} = 900 \, \text{J/kg}^\circ\text{C} \), initial temperature \( T_i = 5^\circ\text{C} \), final temperature \( T_f = 85^\circ\text{C} \). Calculate \( \Delta T = T_f - T_i = 85 - 5 = 80^\circ\text{C} \).

Step3: Substitute into formula

Substitute \( m = 0.10 \), \( c = 900 \), \( \Delta T = 80 \) into \( Q = mc\Delta T \):
\( Q = 0.10 \times 900 \times 80 \).
Calculate \( 0.10 \times 900 = 90 \), then \( 90 \times 80 = 7200 \).

Answer:

7200