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How much heat, in joules and in calories, must be added to a $75.0-\mathrm{g}$ iron block with a specific heat of 0.449 $\mathrm{J} / \mathrm{g}$ $^{\circ} \mathrm{C}$ to increase its temperature from $25^{\circ} \mathrm{C}$ to its melting temperature of $1535^{\circ} \mathrm{C} ?$

50849 joule,12145 cal

01:23

Aadit S.

Chemistry 101

Chapter 5

Thermochemistry

Drexel University

University of Maryland - University College

University of Kentucky

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Rescue calculate the heat required to raise the temperature of the iron block. Former here's Q. Is equal. M. C. Delta T. We're told that the iron block is 75 point several ground. The heat capacity of iron is .449 jules. Program degree Celsius delta T. Final temperature is 15 35. The initial temperature is 25°C solving here we find that this is equal to 50,000 849.3 Jewels. So this would be the amount of heat required and that would be the amount in jewels. Where else to convert this to calories? 50,000 849 3 Jewels. One jewel is equivalent to four calories and this would give us Mhm. Yeah. Mhm. 12,000 yeah. 203.8 calories.

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