How much heat (in kJ) is needed to convert 866 g of ice at -21°C to steam at 126°C? (The specific heats of ice and steam are 2.03 J/g · °C and 1.99 J/g · °C, respectively.)
Added by Aaron M.
Step 1
Q1 = mass * specific heat * change in temperature Q1 = 866 g * 2.03 J/g°C * (0°C - (-21°C)) Q1 = 866 g * 2.03 J/g°C * 21°C Q1 = 36,313.58 J Show more…
Show all steps
Your feedback will help us improve your experience
Sri K and 80 other Chemistry 101 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
How much heat (in $\mathrm{kJ}$ ) is needed to convert $866 \mathrm{g}$ of ice at $-10^{\circ} \mathrm{C}$ to steam at $126^{\circ} \mathrm{C} ?$ (The specific heats of ice and steam are $2.03 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}$ and $1.99 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}$ respectively.)
How much heat (in kilojoules) is needed to convert 212.8 g of ice at -15°C to steam at 138°C? (The specific heat of ice and steam are 2.03 and 1.99 J/g • °C, respectively.)
Amita P.
How much heat (in $\mathrm{kJ} )$ is required to warm 10.0 $\mathrm{g}$ of ice, initially at $-10.0^{\circ} \mathrm{C},$ to steam at $110.0^{\circ} \mathrm{C} ?$ The heat capacity of ice is 2.09 $\mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}$ and that of steam is 2.01 $\mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}$
Recommended Textbooks
Chemistry: Structure and Properties
Chemistry The Central Science
Chemistry
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD