Question

1.750 kg of water is heated to make coco. if it is heated from 20.0°c to 95.0°c, what is the change in thermal energy? (s.h. of water = 4184)
41.84 j
235,350 j
289,110 j
360,870 j
multiple choice 5 points
the specific heat of copper is 385 j/kg•k. if a 2.6 kg block of copper is heated from 300 k to 450 k, how much thermal energy is absorbed?
150,000 j
1000 j
22,000 j

Explanation:

First Question (Water Heating)

Step1: Recall the formula for thermal energy change

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

Step2: Identify the values

Given $m = 0.750\space kg$ (assuming the first number is 0.750, as it's likely a typo), $c = 4184\space J/kg\cdot^\circ C$, initial temperature $T_i = 20.0^\circ C$, final temperature $T_f = 95.0^\circ C$. Calculate $\Delta T = T_f - T_i = 95.0 - 20.0 = 75.0^\circ C$.

Step3: Substitute into the formula

$Q = 0.750\space kg\times4184\space J/kg\cdot^\circ C\times75.0^\circ C$
First, calculate $0.750\times4184 = 3138$, then $3138\times75.0 = 235350\space J$.

Step1: Recall the heat formula

Use $Q = mc\Delta T$, where $m = 2.6\space kg$, $c = 385\space J/kg\cdot K$, $\Delta T = 450 - 300 = 150\space K$ (change in Kelvin is same as change in Celsius).

Step2: Substitute values

$Q = 2.6\space kg\times385\space J/kg\cdot K\times150\space K$
First, $2.6\times385 = 1001$, then $1001\times150 = 150150\space J\approx150,000\space J$.

Answer:

235,350 J

Second Question (Copper Heating)